\displaystyle \textbf{Question 1. }\ \cos^{-1}\!\left(2x\sqrt{1-x^{2}}\right),\ \frac{1}{\sqrt{2}}<x<1
\displaystyle \text{Answer:}

\displaystyle \text{Let, } y=\cos^{-1}\!\left(2x\sqrt{1-x^{2}}\right)

\displaystyle \text{Put } x=\cos\theta

\displaystyle y=\cos^{-1}\!\left(2\cos\theta\sqrt{1-\cos^{2}\theta}\right)

\displaystyle y=\cos^{-1}\!\left(2\cos\theta\sin\theta\right)

\displaystyle y=\cos^{-1}\!\left(\sin2\theta\right)\qquad[\text{Since } \sin2\theta=2\sin\theta\cos\theta]

\displaystyle y=\cos^{-1}\!\left(\cos\left(\frac{\pi}{2}-2\theta\right)\right)\qquad\ldots(i)

\displaystyle \text{Now, } \frac{1}{\sqrt{2}}<x<1

\displaystyle \Rightarrow\ \frac{1}{\sqrt{2}}<\cos\theta<1

\displaystyle \Rightarrow\ 0<\theta<\frac{\pi}{4}

\displaystyle \Rightarrow\ 0<2\theta<\frac{\pi}{2}

\displaystyle \Rightarrow\ 0<\frac{\pi}{2}-2\theta<\frac{\pi}{2}

\displaystyle \text{Hence, from equation }(i)

\displaystyle y=\frac{\pi}{2}-2\theta\qquad[\text{Since } \cos^{-1}(\cos\alpha)=\alpha,\ \alpha\in[0,\pi]]

\displaystyle y=\frac{\pi}{2}-2\cos^{-1}x\qquad[\text{Since } x=\cos\theta]

\displaystyle \text{Differentiating with respect to }x,

\displaystyle \frac{dy}{dx}=\frac{d}{dx}\!\left(\frac{\pi}{2}-2\cos^{-1}x\right)

\displaystyle \frac{dy}{dx}=0-2\left(-\frac{1}{\sqrt{1-x^{2}}}\right)

\displaystyle \frac{dy}{dx}=\frac{2}{\sqrt{1-x^{2}}}

\displaystyle \textbf{Question 2. }\ \cos^{-1}\!\left(\sqrt{\frac{1+x}{2}}\right),\ -1<x<1
\displaystyle \text{Answer:}

\displaystyle \text{Let, } y=\cos^{-1}\!\left(\sqrt{\frac{1+x}{2}}\right)

\displaystyle \text{Put } x=\cos2\theta

\displaystyle y=\cos^{-1}\!\left(\sqrt{\frac{1+\cos2\theta}{2}}\right)

\displaystyle y=\cos^{-1}\!\left(\sqrt{\frac{2\cos^{2}\theta}{2}}\right)

\displaystyle y=\cos^{-1}(\cos\theta)\qquad\ldots(i)

\displaystyle \text{Here, } -1<x<1

\displaystyle \Rightarrow\ -1<\cos2\theta<1

\displaystyle \Rightarrow\ 0<2\theta<\pi

\displaystyle \Rightarrow\ 0<\theta<\frac{\pi}{2}

\displaystyle \text{So, from equation }(i)

\displaystyle y=\theta\qquad[\text{Since } \cos^{-1}(\cos\theta)=\theta,\ \theta\in[0,\pi]]

\displaystyle \Rightarrow\ y=\frac{1}{2}\cos^{-1}x\qquad[\text{Since } x=\cos2\theta]

\displaystyle \text{Differentiating with respect to }x,

\displaystyle \frac{dy}{dx}=\frac{1}{2}\frac{d}{dx}(\cos^{-1}x)

\displaystyle \frac{dy}{dx}=\frac{1}{2}\left(-\frac{1}{\sqrt{1-x^{2}}}\right)

\displaystyle \frac{dy}{dx}=-\frac{1}{2\sqrt{1-x^{2}}}

\displaystyle \textbf{Question 3. }\ \sin^{-1}\!\left(\sqrt{\frac{1-x}{2}}\right),\ 0<x<1
\displaystyle \text{Answer:}

\displaystyle \text{Let, } y=\sin^{-1}\!\left(\sqrt{\frac{1-x}{2}}\right)

\displaystyle \text{Put } x=\cos2\theta

\displaystyle y=\sin^{-1}\!\left(\sqrt{\frac{1-\cos2\theta}{2}}\right)

\displaystyle y=\sin^{-1}\!\left(\sqrt{\frac{2\sin^{2}\theta}{2}}\right)

\displaystyle y=\sin^{-1}(\sin\theta)\qquad\ldots(1)

\displaystyle \text{Here, } 0<x<1

\displaystyle \Rightarrow\ 0<\cos2\theta<1

\displaystyle \Rightarrow\ 0<2\theta<\frac{\pi}{2}

\displaystyle \Rightarrow\ 0<\theta<\frac{\pi}{4}

\displaystyle \text{So, from equation }(1)

\displaystyle y=\theta\qquad[\text{Since } \sin^{-1}(\sin\theta)=\theta,\ \theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]]

\displaystyle \Rightarrow\ y=\frac{1}{2}\cos^{-1}x\qquad[\text{Since } x=\cos2\theta]

\displaystyle \text{Differentiating with respect to }x,

\displaystyle \frac{dy}{dx}=\frac{1}{2}\frac{d}{dx}(\cos^{-1}x)

\displaystyle \frac{dy}{dx}=\frac{1}{2}\left(-\frac{1}{\sqrt{1-x^{2}}}\right)

\displaystyle \frac{dy}{dx}=-\frac{1}{2\sqrt{1-x^{2}}}

\displaystyle \textbf{Question 4. }\ \sin^{-1}\!\left(\sqrt{1-x^{2}}\right),\ 0<x<1
\displaystyle \text{Answer:}

\displaystyle \text{Let, } y=\sin^{-1}\!\left(\sqrt{1-x^{2}}\right)

\displaystyle \text{Put } x=\cos\theta

\displaystyle y=\sin^{-1}\!\left(\sqrt{1-\cos^{2}\theta}\right)

\displaystyle y=\sin^{-1}(\sin\theta)\qquad\ldots(1)

\displaystyle \text{Here, } 0<x<1

\displaystyle \Rightarrow\ 0<\cos\theta<1

\displaystyle \Rightarrow\ 0<\theta<\frac{\pi}{2}

\displaystyle \text{So, from equation }(1)

\displaystyle y=\theta\qquad[\text{Since } \sin^{-1}(\sin\theta)=\theta,\ \theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]]

\displaystyle \Rightarrow\ y=\cos^{-1}x\qquad[\text{Since } x=\cos\theta]

\displaystyle \text{Differentiating with respect to }x,

\displaystyle \frac{dy}{dx}=\frac{d}{dx}(\cos^{-1}x)

\displaystyle \frac{dy}{dx}=-\frac{1}{\sqrt{1-x^{2}}}

\displaystyle \textbf{Question 5. }\ \tan^{-1}\!\left(\frac{x}{\sqrt{a^{2}-x^{2}}}\right),\ -a<x<a
\displaystyle \text{Answer:}

\displaystyle \text{Let, } y=\tan^{-1}\!\left(\frac{x}{\sqrt{a^{2}-x^{2}}}\right)

\displaystyle \text{Put } x=a\sin\theta

\displaystyle y=\tan^{-1}\!\left(\frac{a\sin\theta}{\sqrt{a^{2}-a^{2}\sin^{2}\theta}}\right)

\displaystyle y=\tan^{-1}\!\left(\frac{a\sin\theta}{\sqrt{a^{2}(1-\sin^{2}\theta)}}\right)

\displaystyle y=\tan^{-1}\!\left(\frac{a\sin\theta}{a\cos\theta}\right)

\displaystyle y=\tan^{-1}(\tan\theta)\qquad\ldots(1)

\displaystyle \text{Here, } -a<x<a

\displaystyle \Rightarrow\ -1<\frac{x}{a}<1

\displaystyle \Rightarrow\ -\frac{\pi}{2}<\theta<\frac{\pi}{2}\qquad[\text{Since } x=a\sin\theta]

\displaystyle \text{So, from equation }(1)

\displaystyle y=\theta\qquad[\text{Since } \tan^{-1}(\tan\theta)=\theta,\ \theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)]

\displaystyle \Rightarrow\ y=\sin^{-1}\!\left(\frac{x}{a}\right)\qquad[\text{Since } x=a\sin\theta]

\displaystyle \text{Differentiating with respect to }x,

\displaystyle \text{Using chain rule,}

\displaystyle \frac{dy}{dx}=\frac{1}{\sqrt{1-\left(\frac{x}{a}\right)^{2}}}\cdot\frac{d}{dx}\!\left(\frac{x}{a}\right)

\displaystyle \frac{dy}{dx}=\frac{1}{\sqrt{1-\frac{x^{2}}{a^{2}}}}\cdot\frac{1}{a}

\displaystyle \frac{dy}{dx}=\frac{1}{\sqrt{a^{2}-x^{2}}}

\displaystyle \textbf{Question 6. }\ \sin^{-1}\!\left(\frac{x}{\sqrt{x^{2}+a^{2}}}\right)
\displaystyle \text{Answer:}

\displaystyle \text{Let, } y=\sin^{-1}\!\left(\frac{x}{\sqrt{x^{2}+a^{2}}}\right)

\displaystyle \text{Put } x=a\tan\theta

\displaystyle y=\sin^{-1}\!\left(\frac{a\tan\theta}{\sqrt{a^{2}\tan^{2}\theta+a^{2}}}\right)

\displaystyle y=\sin^{-1}\!\left(\frac{a\tan\theta}{\sqrt{a^{2}(\tan^{2}\theta+1)}}\right)

\displaystyle y=\sin^{-1}\!\left(\frac{a\tan\theta}{a\sec\theta}\right)

\displaystyle y=\sin^{-1}(\sin\theta)

\displaystyle \text{Here, } -\frac{\pi}{2}<\theta<\frac{\pi}{2}

\displaystyle \Rightarrow\ y=\theta\qquad[\text{Since } \sin^{-1}(\sin\theta)=\theta,\ \theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]]

\displaystyle \Rightarrow\ y=\tan^{-1}\!\left(\frac{x}{a}\right)\qquad[\text{Since } x=a\tan\theta]

\displaystyle \text{Differentiating with respect to }x,

\displaystyle \text{Using chain rule,}

\displaystyle \frac{dy}{dx}=\frac{1}{1+\left(\frac{x}{a}\right)^{2}}\cdot\frac{d}{dx}\!\left(\frac{x}{a}\right)

\displaystyle \frac{dy}{dx}=\frac{1}{1+\frac{x^{2}}{a^{2}}}\cdot\frac{1}{a}

\displaystyle \frac{dy}{dx}=\frac{a}{a^{2}+x^{2}}

\displaystyle \textbf{Question 7. }\ \sin^{-1}(2x^{2}-1),\ 0<x<1
\displaystyle \text{Answer:}

\displaystyle \text{Let, } y=\sin^{-1}(2x^{2}-1)

\displaystyle \text{Put } x=\cos\theta

\displaystyle y=\sin^{-1}(2\cos^{2}\theta-1)

\displaystyle y=\sin^{-1}(\cos2\theta)

\displaystyle y=\sin^{-1}\!\left(\sin\!\left(\frac{\pi}{2}-2\theta\right)\right)\qquad\ldots(1)

\displaystyle \text{Here, } 0<x<1

\displaystyle \Rightarrow\ 0<\cos\theta<1

\displaystyle \Rightarrow\ 0<\theta<\frac{\pi}{2}

\displaystyle \Rightarrow\ 0<2\theta<\pi

\displaystyle \Rightarrow\ -\pi<-2\theta<0

\displaystyle \Rightarrow\ -\frac{\pi}{2}<\frac{\pi}{2}-2\theta<\frac{\pi}{2}

\displaystyle \text{So, from equation }(1)

\displaystyle y=\frac{\pi}{2}-2\theta\qquad[\text{Since } \sin^{-1}(\sin\alpha)=\alpha,\ \alpha\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]]

\displaystyle \Rightarrow\ y=\frac{\pi}{2}-2\cos^{-1}x\qquad[\text{Since } x=\cos\theta]

\displaystyle \text{Differentiating with respect to }x,

\displaystyle \frac{dy}{dx}=\frac{d}{dx}\!\left(\frac{\pi}{2}-2\cos^{-1}x\right)

\displaystyle \frac{dy}{dx}=0-2\left(-\frac{1}{\sqrt{1-x^{2}}}\right)

\displaystyle \frac{dy}{dx}=\frac{2}{\sqrt{1-x^{2}}}

\displaystyle \textbf{Question 8. }\ \sin^{-1}(1-2x^{2}),\ 0<x<1
\displaystyle \text{Answer:}

\displaystyle \text{Let, } y=\sin^{-1}(1-2x^{2})

\displaystyle \text{Put } x=\sin\theta

\displaystyle y=\sin^{-1}(1-2\sin^{2}\theta)

\displaystyle y=\sin^{-1}(\cos2\theta)

\displaystyle y=\sin^{-1}\!\left(\sin\!\left(\frac{\pi}{2}-2\theta\right)\right)\qquad\ldots(1)

\displaystyle \text{Here, } 0<x<1

\displaystyle \Rightarrow\ 0<\sin\theta<1

\displaystyle \Rightarrow\ 0<\theta<\frac{\pi}{2}

\displaystyle \Rightarrow\ 0<2\theta<\pi

\displaystyle \Rightarrow\ -\pi<-2\theta<0

\displaystyle \Rightarrow\ -\frac{\pi}{2}<\frac{\pi}{2}-2\theta<\frac{\pi}{2}

\displaystyle \text{So, from equation }(1)

\displaystyle y=\frac{\pi}{2}-2\theta\qquad[\text{Since } \sin^{-1}(\sin\alpha)=\alpha,\ \alpha\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]]

\displaystyle \Rightarrow\ y=\frac{\pi}{2}-2\sin^{-1}x\qquad[\text{Since } x=\sin\theta]

\displaystyle \text{Differentiating with respect to }x,

\displaystyle \frac{dy}{dx}=\frac{d}{dx}\!\left(\frac{\pi}{2}-2\sin^{-1}x\right)

\displaystyle \frac{dy}{dx}=0-2\left(\frac{1}{\sqrt{1-x^{2}}}\right)

\displaystyle \frac{dy}{dx}=-\frac{2}{\sqrt{1-x^{2}}}

\displaystyle \textbf{Question 9. }\ \cos^{-1}\!\left(\frac{x}{\sqrt{x^{2}+a^{2}}}\right)
\displaystyle \text{Answer:}

\displaystyle \text{Let, } y=\cos^{-1}\!\left(\frac{x}{\sqrt{x^{2}+a^{2}}}\right)

\displaystyle \text{Put } x=a\cot\theta

\displaystyle y=\cos^{-1}\!\left(\frac{a\cot\theta}{\sqrt{a^{2}\cot^{2}\theta+a^{2}}}\right)

\displaystyle y=\cos^{-1}\!\left(\frac{a\cot\theta}{\sqrt{a^{2}(\cot^{2}\theta+1)}}\right)

\displaystyle y=\cos^{-1}\!\left(\frac{a\cot\theta}{a\csc\theta}\right)

\displaystyle y=\cos^{-1}(\cos\theta)

\displaystyle \text{Here, } 0<\theta<\pi

\displaystyle \Rightarrow\ y=\theta\qquad[\text{Since } \cos^{-1}(\cos\theta)=\theta,\ \theta\in[0,\pi]]

\displaystyle \Rightarrow\ y=\cot^{-1}\!\left(\frac{x}{a}\right)\qquad[\text{Since } x=a\cot\theta]

\displaystyle \text{Differentiating with respect to }x,

\displaystyle \text{Using chain rule,}

\displaystyle \frac{dy}{dx}=-\frac{1}{1+\left(\frac{x}{a}\right)^{2}}\cdot\frac{d}{dx}\!\left(\frac{x}{a}\right)

\displaystyle \frac{dy}{dx}=-\frac{1}{1+\frac{x^{2}}{a^{2}}}\cdot\frac{1}{a}

\displaystyle \frac{dy}{dx}=-\frac{a}{a^{2}+x^{2}}

\displaystyle \textbf{Question 10. }\ \sin^{-1}\!\left(\frac{\sin x+\cos x}{\sqrt{2}}\right),\ -\frac{3\pi}{4}<x<\frac{\pi}{4}
\displaystyle \text{Answer:}

\displaystyle \text{Let, } y=\sin^{-1}\!\left(\frac{\sin x+\cos x}{\sqrt{2}}\right)

\displaystyle y=\sin^{-1}\!\left(\sin x\cdot\frac{1}{\sqrt{2}}+\cos x\cdot\frac{1}{\sqrt{2}}\right)

\displaystyle y=\sin^{-1}\!\left(\sin x\cos\frac{\pi}{4}+\cos x\sin\frac{\pi}{4}\right)

\displaystyle y=\sin^{-1}\!\left(\sin\!\left(x+\frac{\pi}{4}\right)\right)\qquad\ldots(i)

\displaystyle \text{Here, } -\frac{3\pi}{4}<x<\frac{\pi}{4}

\displaystyle \Rightarrow\ -\frac{3\pi}{4}+\frac{\pi}{4}<x+\frac{\pi}{4}<\frac{\pi}{4}+\frac{\pi}{4}

\displaystyle \Rightarrow\ -\frac{\pi}{2}<x+\frac{\pi}{4}<\frac{\pi}{2}

\displaystyle \text{From }(i)\text{ we get,}

\displaystyle y=x+\frac{\pi}{4}\qquad[\text{Since } \sin^{-1}(\sin\theta)=\theta,\ \theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]]

\displaystyle \text{Differentiating with respect to }x,

\displaystyle \frac{dy}{dx}=\frac{d}{dx}\!\left(x+\frac{\pi}{4}\right)

\displaystyle \frac{dy}{dx}=1+0

\displaystyle \frac{dy}{dx}=1

\displaystyle \textbf{Question 11. }\ \cos^{-1}\!\left(\frac{\cos x+\sin x}{\sqrt{2}}\right),\ -\frac{\pi}{4}<x<\frac{\pi}{4}
\displaystyle \text{Answer:}

\displaystyle \text{Let, } y=\cos^{-1}\!\left(\frac{\cos x+\sin x}{\sqrt{2}}\right)

\displaystyle y=\cos^{-1}\!\left(\frac{1}{\sqrt{2}}\cos x+\frac{1}{\sqrt{2}}\sin x\right)

\displaystyle y=\cos^{-1}\!\left(\cos\frac{\pi}{4}\cos x+\sin\frac{\pi}{4}\sin x\right)

\displaystyle y=\cos^{-1}\!\left(\cos\!\left(\frac{\pi}{4}-x\right)\right)\qquad\ldots(i)

\displaystyle \text{Here, } -\frac{\pi}{4}<x<\frac{\pi}{4}

\displaystyle \Rightarrow\ -\frac{\pi}{4}<\frac{\pi}{4}-x<\frac{3\pi}{4}

\displaystyle \Rightarrow\ 0<\frac{\pi}{4}-x<\frac{\pi}{2}

\displaystyle \text{So, from equation }(i)

\displaystyle y=\frac{\pi}{4}-x\qquad[\text{Since } \cos^{-1}(\cos\theta)=\theta,\ \theta\in[0,\pi]]

\displaystyle \text{Differentiating with respect to }x,

\displaystyle \frac{dy}{dx}=\frac{d}{dx}\!\left(\frac{\pi}{4}-x\right)

\displaystyle \frac{dy}{dx}=0-1

\displaystyle \frac{dy}{dx}=-1

\displaystyle \textbf{Question 12. }\ \tan^{-1}\!\left(\frac{x}{1+\sqrt{1-x^{2}}}\right),\ -1<x<1
\displaystyle \text{Answer:}

\displaystyle \text{Let, } y=\tan^{-1}\!\left(\frac{x}{1+\sqrt{1-x^{2}}}\right)

\displaystyle \text{Put } x=\sin\theta

\displaystyle y=\tan^{-1}\!\left(\frac{\sin\theta}{1+\sqrt{1-\sin^{2}\theta}}\right)

\displaystyle y=\tan^{-1}\!\left(\frac{\sin\theta}{1+\cos\theta}\right)

\displaystyle y=\tan^{-1}\!\left(\frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\cos^{2}\frac{\theta}{2}}\right)

\displaystyle y=\tan^{-1}\!\left(\tan\frac{\theta}{2}\right)\qquad\ldots(i)

\displaystyle \text{Here, } -1<x<1

\displaystyle \Rightarrow\ -1<\sin\theta<1

\displaystyle \Rightarrow\ -\frac{\pi}{2}<\theta<\frac{\pi}{2}

\displaystyle \Rightarrow\ -\frac{\pi}{4}<\frac{\theta}{2}<\frac{\pi}{4}

\displaystyle \text{So, from equation }(i)

\displaystyle y=\frac{\theta}{2}\qquad[\text{Since } \tan^{-1}(\tan\alpha)=\alpha,\ \alpha\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right]]

\displaystyle \Rightarrow\ y=\frac{1}{2}\sin^{-1}x\qquad[\text{Since } x=\sin\theta]

\displaystyle \text{Differentiating with respect to }x,

\displaystyle \frac{dy}{dx}=\frac{1}{2}\frac{d}{dx}(\sin^{-1}x)

\displaystyle \frac{dy}{dx}=\frac{1}{2}\left(\frac{1}{\sqrt{1-x^{2}}}\right)

\displaystyle \frac{dy}{dx}=\frac{1}{2\sqrt{1-x^{2}}}

\displaystyle \textbf{Question 13. }\ \tan^{-1}\!\left(\frac{x}{a+\sqrt{a^{2}-x^{2}}}\right),\ -a<x<a
\displaystyle \text{Answer:}

\displaystyle \text{Let, } y=\tan^{-1}\!\left(\frac{x}{a+\sqrt{a^{2}-x^{2}}}\right)

\displaystyle \text{Put } x=a\sin\theta

\displaystyle y=\tan^{-1}\!\left(\frac{a\sin\theta}{a+\sqrt{a^{2}-a^{2}\sin^{2}\theta}}\right)

\displaystyle y=\tan^{-1}\!\left(\frac{a\sin\theta}{a+\sqrt{a^{2}(1-\sin^{2}\theta)}}\right)

\displaystyle y=\tan^{-1}\!\left(\frac{a\sin\theta}{a+a\cos\theta}\right)

\displaystyle y=\tan^{-1}\!\left(\frac{a\sin\theta}{a(1+\cos\theta)}\right)

\displaystyle y=\tan^{-1}\!\left(\frac{\sin\theta}{1+\cos\theta}\right)

\displaystyle y=\tan^{-1}\!\left(\frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\cos^{2}\frac{\theta}{2}}\right)

\displaystyle y=\tan^{-1}\!\left(\tan\frac{\theta}{2}\right)\qquad\ldots(i)

\displaystyle \text{Here, } -a<x<a

\displaystyle \Rightarrow\ -1<\frac{x}{a}<1

\displaystyle \Rightarrow\ -1<\sin\theta<1

\displaystyle \Rightarrow\ -\frac{\pi}{2}<\theta<\frac{\pi}{2}

\displaystyle \Rightarrow\ -\frac{\pi}{4}<\frac{\theta}{2}<\frac{\pi}{4}

\displaystyle \text{So, from equation }(i)

\displaystyle y=\frac{\theta}{2}\qquad[\text{Since } \tan^{-1}(\tan\alpha)=\alpha,\ \alpha\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)]

\displaystyle \Rightarrow\ y=\frac{1}{2}\sin^{-1}\!\left(\frac{x}{a}\right)\qquad[\text{Since } x=a\sin\theta]

\displaystyle \text{Differentiating with respect to }x,

\displaystyle \frac{dy}{dx}=\frac{1}{2}\cdot\frac{1}{\sqrt{1-\left(\frac{x}{a}\right)^{2}}\!}\cdot\frac{d}{dx}\!\left(\frac{x}{a}\right)

\displaystyle \frac{dy}{dx}=\frac{1}{2}\cdot\frac{1}{\sqrt{1-\frac{x^{2}}{a^{2}}}}\cdot\frac{1}{a}

\displaystyle \frac{dy}{dx}=\frac{1}{2\sqrt{a^{2}-x^{2}}}

\displaystyle \textbf{Question 14. }\ \sin^{-1}\!\left(\frac{x+\sqrt{1-x^{2}}}{\sqrt{2}}\right),\ -1<x<1
\displaystyle \text{Answer:}

\displaystyle \text{Let, } y=\sin^{-1}\!\left(\frac{x+\sqrt{1-x^{2}}}{\sqrt{2}}\right)

\displaystyle \text{Putting } x=\sin\theta

\displaystyle y=\sin^{-1}\!\left(\frac{\sin\theta+\sqrt{1-\sin^{2}\theta}}{\sqrt{2}}\right)

\displaystyle y=\sin^{-1}\!\left(\frac{\sin\theta+\cos\theta}{\sqrt{2}}\right)

\displaystyle y=\sin^{-1}\!\left(\sin\theta\cdot\frac{1}{\sqrt{2}}+\cos\theta\cdot\frac{1}{\sqrt{2}}\right)

\displaystyle y=\sin^{-1}\!\left(\sin\theta\cos\frac{\pi}{4}+\cos\theta\sin\frac{\pi}{4}\right)

\displaystyle y=\sin^{-1}\!\left(\sin\!\left(\theta+\frac{\pi}{4}\right)\right)\qquad\ldots(1)

\displaystyle \text{Here, } -1<x<1

\displaystyle \Rightarrow\ -1<\sin\theta<1

\displaystyle \Rightarrow\ -\frac{\pi}{2}<\theta<\frac{\pi}{2}

\displaystyle \Rightarrow\ -\frac{\pi}{2}+\frac{\pi}{4}<\theta+\frac{\pi}{4}<\frac{\pi}{2}+\frac{\pi}{4}

\displaystyle \Rightarrow\ -\frac{\pi}{4}<\theta+\frac{\pi}{4}<\frac{3\pi}{4}

\displaystyle \text{So, from equation }(1)

\displaystyle y=\theta+\frac{\pi}{4}\qquad[\text{Since } \sin^{-1}(\sin\alpha)=\alpha,\ \alpha\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]]

\displaystyle \Rightarrow\ y=\sin^{-1}x+\frac{\pi}{4}

\displaystyle \text{Differentiating with respect to }x,

\displaystyle \frac{dy}{dx}=\frac{d}{dx}\!\left(\sin^{-1}x+\frac{\pi}{4}\right)

\displaystyle \frac{dy}{dx}=\frac{1}{\sqrt{1-x^{2}}}+0

\displaystyle \frac{dy}{dx}=\frac{1}{\sqrt{1-x^{2}}}

\displaystyle \textbf{Question 15. }\ \cos^{-1}\!\left(\frac{x+\sqrt{1-x^{2}}}{\sqrt{2}}\right),\ -1<x<1
\displaystyle \text{Answer:}

\displaystyle \text{Let, } y=\cos^{-1}\!\left(\frac{x+\sqrt{1-x^{2}}}{\sqrt{2}}\right)

\displaystyle \text{Put } x=\cos\theta

\displaystyle y=\cos^{-1}\!\left(\frac{\cos\theta+\sqrt{1-\cos^{2}\theta}}{\sqrt{2}}\right)

\displaystyle y=\cos^{-1}\!\left(\frac{\cos\theta+\sin\theta}{\sqrt{2}}\right)

\displaystyle y=\cos^{-1}\!\left(\cos\theta\cdot\frac{1}{\sqrt{2}}+\sin\theta\cdot\frac{1}{\sqrt{2}}\right)

\displaystyle y=\cos^{-1}\!\left(\cos\theta\cos\frac{\pi}{4}+\sin\theta\sin\frac{\pi}{4}\right)

\displaystyle y=\cos^{-1}\!\left(\cos\!\left(\theta-\frac{\pi}{4}\right)\right)\qquad\ldots(i)

\displaystyle \text{Here, } -1<x<1

\displaystyle \Rightarrow\ -1<\cos\theta<1

\displaystyle \Rightarrow\ 0<\theta<\pi

\displaystyle \Rightarrow\ -\frac{\pi}{4}<\theta-\frac{\pi}{4}<\frac{3\pi}{4}

\displaystyle \text{So, from equation }(i)

\displaystyle y=\theta-\frac{\pi}{4}\qquad[\text{Since } \cos^{-1}(\cos\alpha)=\alpha,\ \alpha\in[0,\pi]]

\displaystyle \Rightarrow\ y=\cos^{-1}x-\frac{\pi}{4}\qquad[\text{Since } x=\cos\theta]

\displaystyle \text{Differentiating with respect to }x,

\displaystyle \frac{dy}{dx}=\frac{d}{dx}\!\left(\cos^{-1}x-\frac{\pi}{4}\right)

\displaystyle \frac{dy}{dx}=-\frac{1}{\sqrt{1-x^{2}}}+0

\displaystyle \frac{dy}{dx}=-\frac{1}{\sqrt{1-x^{2}}}

\displaystyle \textbf{Question 16. }\ \tan^{-1}\!\left(\frac{4x}{1-4x^{2}}\right),\ -\frac{1}{2}<x<\frac{1}{2}
\displaystyle \text{Answer:}

\displaystyle \text{Let, } y=\tan^{-1}\!\left(\frac{4x}{1-4x^{2}}\right)

\displaystyle \text{Put } 2x=\tan\theta

\displaystyle y=\tan^{-1}\!\left(\frac{2\tan\theta}{1-\tan^{2}\theta}\right)

\displaystyle y=\tan^{-1}(\tan2\theta)\qquad\ldots(i)

\displaystyle \text{Here, } -\frac{1}{2}<x<\frac{1}{2}

\displaystyle \Rightarrow\ -1<2x<1

\displaystyle \Rightarrow\ -1<\tan\theta<1

\displaystyle \Rightarrow\ -\frac{\pi}{4}<\theta<\frac{\pi}{4}

\displaystyle \Rightarrow\ -\frac{\pi}{2}<2\theta<\frac{\pi}{2}

\displaystyle \text{So, from equation }(i)

\displaystyle y=2\theta\qquad[\text{Since } \tan^{-1}(\tan\alpha)=\alpha,\ \alpha\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)]

\displaystyle \Rightarrow\ y=2\tan^{-1}(2x)\qquad[\text{Since } 2x=\tan\theta]

\displaystyle \text{Differentiating with respect to }x,

\displaystyle \frac{dy}{dx}=2\cdot\frac{1}{1+(2x)^{2}}\cdot\frac{d}{dx}(2x)

\displaystyle \frac{dy}{dx}=2\cdot\frac{1}{1+4x^{2}}\cdot2

\displaystyle \frac{dy}{dx}=\frac{4}{1+4x^{2}}

\displaystyle \textbf{Question 17. }\ \tan^{-1}\!\left(\frac{2^{x}+1}{1-4^{x}}\right),\ -\infty<x<0
\displaystyle \text{Answer:}

\displaystyle \text{Let, } y=\tan^{-1}\!\left(\frac{2^{x+1}}{1-4^{x}}\right)

\displaystyle \text{Put } 2^{x}=\tan\theta

\displaystyle y=\tan^{-1}\!\left(\frac{2^{x}\cdot 2}{1-(2^{x})^{2}}\right)

\displaystyle y=\tan^{-1}\!\left(\frac{2\tan\theta}{1-\tan^{2}\theta}\right)

\displaystyle y=\tan^{-1}(\tan2\theta)\qquad\ldots(i)

\displaystyle \text{Here, } -\infty<x<0

\displaystyle \Rightarrow\ 0<2^{x}<1

\displaystyle \Rightarrow\ 0<\tan\theta<1

\displaystyle \Rightarrow\ 0<\theta<\frac{\pi}{4}

\displaystyle \Rightarrow\ 0<2\theta<\frac{\pi}{2}

\displaystyle \text{So, from equation }(i)

\displaystyle y=2\theta\qquad[\text{Since } \tan^{-1}(\tan\alpha)=\alpha,\ \alpha\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)]

\displaystyle \Rightarrow\ y=2\tan^{-1}(2^{x})

\displaystyle \text{Differentiating with respect to }x,

\displaystyle \frac{dy}{dx}=2\cdot\frac{1}{1+(2^{x})^{2}}\cdot\frac{d}{dx}(2^{x})

\displaystyle \frac{dy}{dx}=\frac{2\cdot 2^{x}\ln2}{1+4^{x}}

\displaystyle \frac{dy}{dx}=\frac{2^{x+1}\ln2}{1+4^{x}}

\displaystyle \textbf{Question 18. }\ \tan^{-1}\!\left(\frac{2a^{x}}{1-a^{2x}}\right),\ a>1,\ -\infty<x<0
\displaystyle \text{Answer:}

\displaystyle \text{Let, } y=\tan^{-1}\!\left(\frac{2a^{x}}{1-a^{2x}}\right)

\displaystyle \text{Put } a^{x}=\tan\theta

\displaystyle y=\tan^{-1}\!\left(\frac{2\tan\theta}{1-\tan^{2}\theta}\right)

\displaystyle y=\tan^{-1}(\tan2\theta)\qquad\ldots(i)

\displaystyle \text{Here, } -\infty<x<0

\displaystyle \Rightarrow\ 0<a^{x}<1\qquad(a>1)

\displaystyle \Rightarrow\ 0<\tan\theta<1

\displaystyle \Rightarrow\ 0<\theta<\frac{\pi}{4}

\displaystyle \Rightarrow\ 0<2\theta<\frac{\pi}{2}

\displaystyle \text{So, from equation }(i)

\displaystyle y=2\theta\qquad[\text{Since } \tan^{-1}(\tan\alpha)=\alpha,\ \alpha\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)]

\displaystyle \Rightarrow\ y=2\tan^{-1}(a^{x})

\displaystyle \text{Differentiating with respect to }x,

\displaystyle \frac{dy}{dx}=2\cdot\frac{1}{1+(a^{x})^{2}}\cdot\frac{d}{dx}(a^{x})

\displaystyle \frac{dy}{dx}=\frac{2a^{x}\ln a}{1+a^{2x}}

\displaystyle \textbf{Question 19. }\ \sin^{-1}\!\left(\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right),\ 0<x<1
\displaystyle \text{Answer:}

\displaystyle \text{Let, } y=\sin^{-1}\!\left(\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right)

\displaystyle \text{Put } x=\cos2\theta

\displaystyle y=\sin^{-1}\!\left(\frac{\sqrt{1+\cos2\theta}+\sqrt{1-\cos2\theta}}{2}\right)

\displaystyle y=\sin^{-1}\!\left(\frac{\sqrt{2\cos^{2}\theta}+\sqrt{2\sin^{2}\theta}}{2}\right)

\displaystyle y=\sin^{-1}\!\left(\frac{\sqrt{2}\cos\theta+\sqrt{2}\sin\theta}{2}\right)

\displaystyle y=\sin^{-1}\!\left(\cos\theta\cdot\frac{1}{\sqrt{2}}+\sin\theta\cdot\frac{1}{\sqrt{2}}\right)

\displaystyle y=\sin^{-1}\!\left(\cos\theta\sin\frac{\pi}{4}+\sin\theta\cos\frac{\pi}{4}\right)

\displaystyle y=\sin^{-1}\!\left(\sin\!\left(\theta+\frac{\pi}{4}\right)\right)\qquad\ldots(i)

\displaystyle \text{Here, } 0<x<1

\displaystyle \Rightarrow\ 0<\cos2\theta<1

\displaystyle \Rightarrow\ 0<2\theta<\frac{\pi}{2}

\displaystyle \Rightarrow\ 0<\theta<\frac{\pi}{4}

\displaystyle \Rightarrow\ \frac{\pi}{4}<\theta+\frac{\pi}{4}<\frac{\pi}{2}

\displaystyle \text{So, from equation }(i)

\displaystyle y=\theta+\frac{\pi}{4}\qquad[\text{Since } \sin^{-1}(\sin\alpha)=\alpha,\ \alpha\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]]

\displaystyle \Rightarrow\ y=\frac{1}{2}\cos^{-1}x+\frac{\pi}{4}\qquad[\text{Since } x=\cos2\theta]

\displaystyle \text{Differentiating with respect to }x,

\displaystyle \frac{dy}{dx}=\frac{1}{2}\frac{d}{dx}(\cos^{-1}x)+0

\displaystyle \frac{dy}{dx}=\frac{1}{2}\left(-\frac{1}{\sqrt{1-x^{2}}}\right)

\displaystyle \frac{dy}{dx}=-\frac{1}{2\sqrt{1-x^{2}}}

\displaystyle \textbf{Question 20. }\ \tan^{-1}\!\left(\frac{\sqrt{1+a^{2}x^{2}}-1}{ax}\right),\ x\ne 0
\displaystyle \text{Answer:}

\displaystyle \text{Let, } y=\tan^{-1}\!\left(\frac{\sqrt{1+a^{2}x^{2}}-1}{ax}\right)

\displaystyle \text{Put } ax=\tan\theta

\displaystyle y=\tan^{-1}\!\left(\frac{\sqrt{1+\tan^{2}\theta}-1}{\tan\theta}\right)

\displaystyle y=\tan^{-1}\!\left(\frac{\sec\theta-1}{\tan\theta}\right)

\displaystyle y=\tan^{-1}\!\left(\frac{1-\cos\theta}{\sin\theta}\right)

\displaystyle y=\tan^{-1}\!\left(\frac{2\sin^{2}\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}\right)

\displaystyle y=\tan^{-1}\!\left(\tan\frac{\theta}{2}\right)

\displaystyle y=\frac{\theta}{2}\qquad[\text{Since } \tan^{-1}(\tan\alpha)=\alpha,\ \alpha\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)]

\displaystyle \Rightarrow\ y=\frac{1}{2}\tan^{-1}(ax)

\displaystyle \text{Differentiating with respect to }x,

\displaystyle \frac{dy}{dx}=\frac{1}{2}\cdot\frac{1}{1+(ax)^{2}}\cdot\frac{d}{dx}(ax)

\displaystyle \frac{dy}{dx}=\frac{1}{2}\cdot\frac{1}{1+a^{2}x^{2}}\cdot a

\displaystyle \frac{dy}{dx}=\frac{a}{2(1+a^{2}x^{2})}

\displaystyle \textbf{Question 21. }\ \tan^{-1}\!\left(\frac{\sin x}{1+\cos x}\right),\ -\pi<x<\pi
\displaystyle \text{Answer:}

\displaystyle \text{Let } f(x)=\tan^{-1}\!\left(\frac{\sin x}{1+\cos x}\right)

\displaystyle \text{This function is defined for all real }x\text{ where }1+\cos x\neq 0

\displaystyle f(x)=\tan^{-1}\!\left(\frac{2\sin\frac{x}{2}\cos\frac{x}{2}}{2\cos^{2}\frac{x}{2}}\right)

\displaystyle f(x)=\tan^{-1}\!\left(\tan\frac{x}{2}\right)

\displaystyle f(x)=\frac{x}{2}\qquad[\text{Since } \tan^{-1}(\tan\theta)=\theta,\ \theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)]

\displaystyle \text{Thus, } f'(x)=\frac{d}{dx}\!\left(\frac{x}{2}\right)

\displaystyle f'(x)=\frac{1}{2}

\displaystyle \textbf{Question 22. }\ \sin^{-1}\!\left(\frac{1}{\sqrt{1+x^{2}}}\right)
\displaystyle \text{Answer:}

\displaystyle \text{Let, } y=\sin^{-1}\!\left(\frac{1}{\sqrt{1+x^{2}}}\right)

\displaystyle \text{Put } x=\cot\theta

\displaystyle y=\sin^{-1}\!\left(\frac{1}{\sqrt{1+\cot^{2}\theta}}\right)

\displaystyle y=\sin^{-1}\!\left(\frac{1}{\sqrt{\mathrm{cosec}^{2}\theta}}\right)

\displaystyle y=\sin^{-1}(\sin\theta)

\displaystyle \text{Here, } 0<\theta<\frac{\pi}{2}

\displaystyle \Rightarrow\ y=\theta\qquad[\text{Since } \sin^{-1}(\sin\theta)=\theta,\ \theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]]

\displaystyle \Rightarrow\ y=\cot^{-1}x\qquad[\text{Since } x=\cot\theta]

\displaystyle \text{Differentiating with respect to }x,

\displaystyle \frac{dy}{dx}=\frac{d}{dx}(\cot^{-1}x)

\displaystyle \frac{dy}{dx}=-\frac{1}{1+x^{2}}

\displaystyle \textbf{Question 23. }\ \cos^{-1}\!\left(\frac{1-x^{2n}}{1+x^{2n}}\right),\ 0<x<\infty
\displaystyle \text{Answer:}

\displaystyle \text{Let, } y=\cos^{-1}\!\left(\frac{1-x^{2n}}{1+x^{2n}}\right)

\displaystyle \text{Put } x^{n}=\tan\theta

\displaystyle y=\cos^{-1}\!\left(\frac{1-(x^{n})^{2}}{1+(x^{n})^{2}}\right)

\displaystyle y=\cos^{-1}\!\left(\frac{1-\tan^{2}\theta}{1+\tan^{2}\theta}\right)

\displaystyle y=\cos^{-1}(\cos2\theta)\qquad\ldots(i)

\displaystyle \text{Here, } 0<x<\infty

\displaystyle \Rightarrow\ 0<x^{n}<\infty

\displaystyle \Rightarrow\ 0<\tan\theta<\infty

\displaystyle \Rightarrow\ 0<\theta<\frac{\pi}{2}

\displaystyle \Rightarrow\ 0<2\theta<\pi

\displaystyle \text{So, from equation }(i)

\displaystyle y=2\theta\qquad[\text{Since } \cos^{-1}(\cos\alpha)=\alpha,\ \alpha\in[0,\pi]]

\displaystyle \Rightarrow\ y=2\tan^{-1}(x^{n})

\displaystyle \text{Differentiating with respect to }x\text{ using chain rule,}

\displaystyle \frac{dy}{dx}=2\left[\frac{1}{1+(x^{n})^{2}}\right]\frac{d}{dx}(x^{n})

\displaystyle \frac{dy}{dx}=2\left(\frac{1}{1+x^{2n}}\right)(nx^{n-1})

\displaystyle \frac{dy}{dx}=\frac{2nx^{\,n-1}}{1+x^{2n}}

\displaystyle \textbf{Question 24. }\ \sin^{-1}\!\left(\frac{1-x^{2}}{1+x^{2}}\right)+\sec^{-1}\!\left(\frac{1+x^{2}}{1-x^{2}}\right),\ x\in R
\displaystyle \text{Answer:}

\displaystyle \text{Let, } y=\sin^{-1}\!\left(\frac{1-x^{2}}{1+x^{2}}\right)+\sec^{-1}\!\left(\frac{1+x^{2}}{1-x^{2}}\right)

\displaystyle y=\sin^{-1}\!\left(\frac{1-x^{2}}{1+x^{2}}\right)+\cos^{-1}\!\left(\frac{1-x^{2}}{1+x^{2}}\right) \qquad[\text{Since } \sec^{-1}x=\cos^{-1}\!\left(\frac{1}{x}\right)]

\displaystyle y=\frac{\pi}{2}\qquad[\text{Since } \sin^{-1}u+\cos^{-1}u=\frac{\pi}{2}]

\displaystyle \text{Differentiating with respect to }x,

\displaystyle \frac{dy}{dx}=0

\displaystyle \textbf{Question 25. }\ \tan^{-1}\!\left(\frac{a+x}{1-ax}\right)
\displaystyle \text{Answer:}

\displaystyle \text{Let, } y=\tan^{-1}\!\left(\frac{a+x}{1-ax}\right)

\displaystyle y=\tan^{-1}a+\tan^{-1}x \qquad[\text{Since } \tan^{-1}u+\tan^{-1}v=\tan^{-1}\!\left(\frac{u+v}{1-uv}\right)]

\displaystyle \text{Differentiating with respect to }x,

\displaystyle \frac{dy}{dx}=\frac{d}{dx}(\tan^{-1}a)+\frac{d}{dx}(\tan^{-1}x)

\displaystyle \frac{dy}{dx}=0+\frac{1}{1+x^{2}}

\displaystyle \frac{dy}{dx}=\frac{1}{1+x^{2}}

\displaystyle \textbf{Question 26. }\ \tan^{-1}\!\left(\frac{\sqrt{x}+\sqrt{a}}{1-\sqrt{xa}}\right)
\displaystyle \text{Answer:}

\displaystyle \text{Let, } y=\tan^{-1}\!\left(\frac{\sqrt{x}+\sqrt{a}}{1-\sqrt{ax}}\right)

\displaystyle y=\tan^{-1}(\sqrt{x})+\tan^{-1}(\sqrt{a}) \qquad[\text{Since } \tan^{-1}u+\tan^{-1}v=\tan^{-1}\!\left(\frac{u+v}{1-uv}\right)]

\displaystyle \text{Differentiating with respect to }x\text{ using chain rule,}

\displaystyle \frac{dy}{dx}=\frac{d}{dx}(\tan^{-1}\sqrt{x})+\frac{d}{dx}(\tan^{-1}\sqrt{a})

\displaystyle \frac{dy}{dx}=\frac{1}{1+(\sqrt{x})^{2}}\cdot\frac{d}{dx}(\sqrt{x})+0

\displaystyle \frac{dy}{dx}=\frac{1}{1+x}\cdot\frac{1}{2\sqrt{x}}

\displaystyle \frac{dy}{dx}=\frac{1}{2\sqrt{x}(1+x)}

\displaystyle \textbf{Question 27. }\ \tan^{-1}\!\left(\frac{a+b\tan x}{b-a\tan x}\right)
\displaystyle \text{Answer:}

\displaystyle \text{Let, } y=\tan^{-1}\!\left(\frac{a+b\tan x}{\,b-a\tan x\,}\right)

\displaystyle y=\tan^{-1}\!\left(\frac{\frac{a}{b}+\tan x}{1-\frac{a}{b}\tan x}\right)

\displaystyle y=\tan^{-1}\!\left( \frac{\tan\!\left(\tan^{-1}\frac{a}{b}\right)+\tan x} {1-\tan\!\left(\tan^{-1}\frac{a}{b}\right)\tan x} \right)

\displaystyle y=\tan^{-1}\!\left(\tan\!\left(\tan^{-1}\frac{a}{b}+x\right)\right)

\displaystyle y=\tan^{-1}\!\left(\frac{a}{b}\right)+x \qquad[\text{Since } \tan^{-1}(\tan\theta)=\theta,\ \theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)]

\displaystyle \text{Differentiating with respect to }x,

\displaystyle \frac{dy}{dx}=\frac{d}{dx}\!\left(\tan^{-1}\!\frac{a}{b}\right)+\frac{d}{dx}(x)

\displaystyle \frac{dy}{dx}=0+1

\displaystyle \frac{dy}{dx}=1

\displaystyle \textbf{Question 28. }\ \tan^{-1}\!\left(\frac{a+bx}{b-ax}\right)
\displaystyle \text{Answer:}

\displaystyle \text{Let, } y=\tan^{-1}\!\left(\frac{a+bx}{b-ax}\right)

\displaystyle y=\tan^{-1}\!\left(\frac{\frac{a}{b}+x}{1-\frac{a}{b}x}\right)

\displaystyle y=\tan^{-1}\!\left( \frac{\tan\!\left(\tan^{-1}\frac{a}{b}\right)+\tan(\tan^{-1}x)} {1-\tan\!\left(\tan^{-1}\frac{a}{b}\right)\tan(\tan^{-1}x)} \right)

\displaystyle y=\tan^{-1}\!\left(\tan\!\left(\tan^{-1}\frac{a}{b}+\tan^{-1}x\right)\right)

\displaystyle y=\tan^{-1}\!\left(\frac{a}{b}\right)+\tan^{-1}x \qquad[\text{Since } \tan^{-1}(\tan\theta)=\theta,\ \theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)]

\displaystyle \text{Differentiating with respect to }x,

\displaystyle \frac{dy}{dx} =\frac{d}{dx}\!\left(\tan^{-1}\!\frac{a}{b}\right) +\frac{d}{dx}(\tan^{-1}x)

\displaystyle \frac{dy}{dx}=0+\frac{1}{1+x^{2}}

\displaystyle \frac{dy}{dx}=\frac{1}{1+x^{2}}

\displaystyle \textbf{Question 29. }\ \tan^{-1}\!\left(\frac{x-a}{x+a}\right)
\displaystyle \text{Answer:}

\displaystyle \text{Let, } y=\tan^{-1}\!\left(\frac{x-a}{x+a}\right)

\displaystyle y=\tan^{-1}\!\left(\frac{\frac{x}{x}-\frac{a}{x}}{\frac{x}{x}+\frac{a}{x}}\right)

\displaystyle y=\tan^{-1}\!\left(\frac{1-\frac{a}{x}}{1+\frac{a}{x}}\right)

\displaystyle y=\tan^{-1}(1)-\tan^{-1}\!\left(\frac{a}{x}\right) \qquad[\text{Since } \tan^{-1}u-\tan^{-1}v=\tan^{-1}\!\left(\frac{u-v}{1+uv}\right)]

\displaystyle \text{Differentiating with respect to }x,

\displaystyle \frac{dy}{dx} =0-\frac{1}{1+\left(\frac{a}{x}\right)^{2}}\cdot\frac{d}{dx}\!\left(\frac{a}{x}\right)

\displaystyle \frac{dy}{dx} =-\frac{1}{1+\frac{a^{2}}{x^{2}}}\cdot\left(-\frac{a}{x^{2}}\right)

\displaystyle \frac{dy}{dx} =\frac{a}{x^{2}+a^{2}}

\displaystyle \textbf{Question 30. }\ \tan^{-1}\!\left(\frac{x}{1+6x^{2}}\right)
\displaystyle \text{Answer:}

\displaystyle \text{Let, } y=\tan^{-1}\!\left(\frac{x}{1+6x^{2}}\right)

\displaystyle y=\tan^{-1}\!\left(\frac{3x-2x}{1+(3x)(2x)}\right)

\displaystyle y=\tan^{-1}(3x)-\tan^{-1}(2x) \qquad[\text{Since } \tan^{-1}u-\tan^{-1}v=\tan^{-1}\!\left(\frac{u-v}{1+uv}\right)]

\displaystyle \text{Differentiating with respect to }x\text{ using chain rule,}

\displaystyle \frac{dy}{dx} =\frac{1}{1+(3x)^{2}}\frac{d}{dx}(3x) -\frac{1}{1+(2x)^{2}}\frac{d}{dx}(2x)

\displaystyle \frac{dy}{dx} =\frac{3}{1+9x^{2}}-\frac{2}{1+4x^{2}}

\displaystyle \textbf{Question 31. }\ \tan^{-1}\!\left(\frac{5x}{1-6x^{2}}\right),\ -\frac{1}{\sqrt{6}}<x<\frac{1}{\sqrt{6}}
\displaystyle \text{Answer:}

\displaystyle \text{Let, } y=\tan^{-1}\!\left(\frac{5x}{1-6x^{2}}\right)

\displaystyle y=\tan^{-1}\!\left(\frac{3x+2x}{1-(3x)(2x)}\right)

\displaystyle y=\tan^{-1}(3x)+\tan^{-1}(2x) \qquad[\text{Since } \tan^{-1}u+\tan^{-1}v=\tan^{-1}\!\left(\frac{u+v}{1-uv}\right)]

\displaystyle \text{Differentiating with respect to }x\text{ using chain rule,}

\displaystyle \frac{dy}{dx} =\frac{1}{1+(3x)^{2}}\frac{d}{dx}(3x) +\frac{1}{1+(2x)^{2}}\frac{d}{dx}(2x)

\displaystyle \frac{dy}{dx} =\frac{3}{1+9x^{2}}+\frac{2}{1+4x^{2}}

\displaystyle \textbf{Question 32. }\ \tan^{-1}\!\left(\frac{\cos x+\sin x}{\cos x-\sin x}\right),\ -\frac{\pi}{4}<x<\frac{\pi}{4}
\displaystyle \text{Answer:}

\displaystyle \text{Let, } y=\tan^{-1}\!\left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)

\displaystyle y=\tan^{-1}\!\left( \frac{\frac{\cos x}{\cos x}+\frac{\sin x}{\cos x}} {\frac{\cos x}{\cos x}-\frac{\sin x}{\cos x}} \right)

\displaystyle y=\tan^{-1}\!\left(\frac{1+\tan x}{1-\tan x}\right)

\displaystyle y=\tan^{-1}\!\left( \frac{\tan\frac{\pi}{4}+\tan x}{1-\tan\frac{\pi}{4}\tan x} \right)

\displaystyle y=\tan^{-1}\!\left(\tan\!\left(\frac{\pi}{4}+x\right)\right)

\displaystyle y=\frac{\pi}{4}+x \qquad[\text{Since } \tan^{-1}(\tan\theta)=\theta,\ \theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)]

\displaystyle \text{Differentiating with respect to }x,

\displaystyle \frac{dy}{dx}=\frac{d}{dx}\!\left(\frac{\pi}{4}+x\right)

\displaystyle \frac{dy}{dx}=0+1

\displaystyle \frac{dy}{dx}=1

\displaystyle \textbf{Question 33. }\ \tan^{-1}\!\left(\frac{x^{1/3}+a^{1/3}}{1-(ax)^{1/3}}\right)
\displaystyle \text{Answer:}

\displaystyle \text{Let, } y=\tan^{-1}\!\left(\frac{x^{\frac13}+a^{\frac13}}{1-(ax)^{\frac13}}\right)

\displaystyle y=\tan^{-1}\!\left(x^{\frac13}\right)+\tan^{-1}\!\left(a^{\frac13}\right) \qquad[\text{Since } \tan^{-1}u+\tan^{-1}v=\tan^{-1}\!\left(\frac{u+v}{1-uv}\right)]

\displaystyle \text{Differentiating with respect to }x\text{ using chain rule,}

\displaystyle \frac{dy}{dx} =\frac{1}{1+\left(x^{\frac13}\right)^{2}}\cdot\frac{d}{dx}\!\left(x^{\frac13}\right)+0

\displaystyle \frac{dy}{dx} =\frac{1}{1+x^{\frac23}}\cdot\frac{1}{3}x^{-\frac23}

\displaystyle \frac{dy}{dx} =\frac{1}{3x^{\frac23}\left(1+x^{\frac23}\right)}

\displaystyle \textbf{Question 34. }\ \sin^{-1}\!\left(\frac{2^{x+1}}{1+4^{x}}\right)
\displaystyle \text{Answer:}

\displaystyle \text{Let } y=\sin^{-1}\!\left(\frac{1}{\sqrt{1+x^{2}}}\right)

\displaystyle \text{Putting } x=\cot\theta \Rightarrow \theta=\cot^{-1}x

\displaystyle y=\sin^{-1}\!\left(\frac{1}{\sqrt{1+(\cot\theta)^{2}}}\right)

\displaystyle y=\sin^{-1}\!\left(\frac{1}{\sqrt{\mathrm{cosec}^{2}\theta}}\right)

\displaystyle y=\sin^{-1}(\sin\theta)

\displaystyle y=\theta \qquad[\text{Since } \sin^{-1}(\sin\theta)=\theta,\ \theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]]

\displaystyle \therefore\ y=\cot^{-1}x

\displaystyle \text{Differentiating with respect to }x,

\displaystyle \frac{dy}{dx}=\frac{d}{dx}(\cot^{-1}x)

\displaystyle \frac{dy}{dx}=-\frac{1}{1+x^{2}}

\displaystyle \textbf{Question 35. }\ \text{If } y=\sin^{-1}\!\left(\frac{2x}{1+x^{2}}\right)+\sec^{-1}\!\left(\frac{1+x^{2}}{1-x^{2}}\right),\ 0<x<1,\ \text{prove that }\frac{dy}{dx}=\frac{4}{1+x^{2}}
\displaystyle \text{Answer:}

\displaystyle \text{Let, } y=\sin^{-1}\!\left(\frac{2x}{1+x^{2}}\right) +\sec^{-1}\!\left(\frac{1+x^{2}}{1-x^{2}}\right)

\displaystyle y=\sin^{-1}\!\left(\frac{2x}{1+x^{2}}\right) +\cos^{-1}\!\left(\frac{1-x^{2}}{1+x^{2}}\right) \qquad[\text{Since } \sec^{-1}u=\cos^{-1}\!\left(\frac{1}{u}\right)]

\displaystyle \text{Put } x=\tan\theta

\displaystyle y=\sin^{-1}\!\left(\frac{2\tan\theta}{1+\tan^{2}\theta}\right) +\cos^{-1}\!\left(\frac{1-\tan^{2}\theta}{1+\tan^{2}\theta}\right)

\displaystyle y=\sin^{-1}(\sin2\theta)+\cos^{-1}(\cos2\theta)\qquad\ldots(i)

\displaystyle \text{Here, } 0<x<1

\displaystyle \Rightarrow\ 0<\tan\theta<1

\displaystyle \Rightarrow\ 0<\theta<\frac{\pi}{4}

\displaystyle \Rightarrow\ 0<2\theta<\frac{\pi}{2}

\displaystyle \text{So, from equation }(i)

\displaystyle y=2\theta+2\theta \quad[\text{Since } \sin^{-1}(\sin\alpha)=\alpha,\ \alpha\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right],\ \cos^{-1}(\cos\beta)=\beta,\ \beta\in[0,\pi]]

\displaystyle y=4\theta

\displaystyle \Rightarrow\ y=4\tan^{-1}x\qquad[\text{Since } x=\tan\theta]

\displaystyle \text{Differentiating with respect to }x,

\displaystyle \frac{dy}{dx} =4\cdot\frac{1}{1+x^{2}}

\displaystyle \frac{dy}{dx}=\frac{4}{1+x^{2}}

\displaystyle \textbf{Question 36. }\ \text{If } y=\sin^{-1}\!\left(\frac{x}{\sqrt{1+x^{2}}}\right)+\cos^{-1}\!\left(\frac{1}{\sqrt{1+x^{2}}}\right),\ 0<x<\infty,\ \text{prove that }\frac{dy}{dx}=\frac{2}{1+x^{2}}
\displaystyle \text{Answer:}

\displaystyle \text{Let, } y=\sin^{-1}\!\left(\frac{x}{\sqrt{1+x^{2}}}\right) +\cos^{-1}\!\left(\frac{1}{\sqrt{1+x^{2}}}\right)

\displaystyle \text{Put } x=\tan\theta

\displaystyle y=\sin^{-1}\!\left(\frac{\tan\theta}{\sqrt{1+\tan^{2}\theta}}\right) +\cos^{-1}\!\left(\frac{1}{\sqrt{1+\tan^{2}\theta}}\right)

\displaystyle y=\sin^{-1}\!\left(\frac{\sin\theta}{\cos\theta}\cdot\cos\theta\right) +\cos^{-1}(\cos\theta)

\displaystyle y=\sin^{-1}(\sin\theta)+\cos^{-1}(\cos\theta)\qquad\ldots(i)

\displaystyle \text{Here, } 0<x<\infty

\displaystyle \Rightarrow\ 0<\tan\theta<\infty

\displaystyle \Rightarrow\ 0<\theta<\frac{\pi}{2}

\displaystyle \text{So, from equation }(i)

\displaystyle y=\theta+\theta \quad[\text{Since } \sin^{-1}(\sin\alpha)=\alpha,\ \alpha\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right],\ \cos^{-1}(\cos\beta)=\beta,\ \beta\in[0,\pi]]

\displaystyle y=2\theta

\displaystyle \Rightarrow\ y=2\tan^{-1}x\qquad[\text{Since } x=\tan\theta]

\displaystyle \text{Differentiating with respect to }x,

\displaystyle \frac{dy}{dx} =2\cdot\frac{1}{1+x^{2}}

\displaystyle \frac{dy}{dx}=\frac{2}{1+x^{2}}

\displaystyle \textbf{Question 37. }\ \text{Differentiate the following with respect to }x:

\displaystyle \textbf{(i) } \cos^{-1}( \sin x)   \ \ \ \ \ \  \textbf{(ii) } \ \cot^{-1}\!\left(\frac{1-x}{1+x}\right)
\displaystyle \text{Answer:}

\displaystyle \text{(i). }

\displaystyle \text{Let } f(x)=\cos^{-1}(\sin x)

\displaystyle f(x)=\cos^{-1}\!\left(\cos\!\left(\frac{\pi}{2}-x\right)\right)

\displaystyle \text{Here, } -\frac{\pi}{2}\le \frac{\pi}{2}-x \le \pi

\displaystyle \Rightarrow\ f(x)=\frac{\pi}{2}-x \qquad[\text{Since } \cos^{-1}(\cos\theta)=\theta,\ \theta\in[0,\pi]]

\displaystyle \text{Thus, } f'(x)=\frac{d}{dx}\!\left(\frac{\pi}{2}-x\right)

\displaystyle f'(x)=-1

\displaystyle \text{(ii). }

\displaystyle \frac{d}{dx}\!\left[\cot^{-1}\!\left(\frac{1-x}{1+x}\right)\right] =\frac{d}{dx}\!\left[\tan^{-1}\!\left(\frac{1+x}{1-x}\right)\right]

\displaystyle =\frac{1}{1+\left(\frac{1+x}{1-x}\right)^{2}} \cdot\frac{d}{dx}\!\left(\frac{1+x}{1-x}\right)

\displaystyle =\frac{1}{1+\left(\frac{1+x}{1-x}\right)^{2}} \cdot\frac{(1-x)+(1+x)}{(1-x)^{2}}

\displaystyle =\frac{(1-x)^{2}}{(1-x)^{2}+(1+x)^{2}}\cdot\frac{2}{(1-x)^{2}}

\displaystyle =\frac{2}{(1-x)^{2}+(1+x)^{2}}

\displaystyle =\frac{2}{2+2x^{2}}

\displaystyle =\frac{1}{1+x^{2}}

\displaystyle \textbf{Question 38. }\ \text{If } y=\cot^{-1}\!\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right),\ \text{show that }\frac{dy}{dx}\ \text{is independent of }x
\displaystyle \text{Answer:}

\displaystyle \text{Let, } y=\cot^{-1}\!\left( \frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}} {\sqrt{1+\sin x}-\sqrt{1-\sin x}} \right)\qquad\ldots(i)

\displaystyle \text{We have, }

\displaystyle \frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}} {\sqrt{1+\sin x}-\sqrt{1-\sin x}} =\frac{\left(\sqrt{1+\sin x}+\sqrt{1-\sin x}\right)^{2}} {\left(\sqrt{1+\sin x}-\sqrt{1-\sin x}\right) \left(\sqrt{1+\sin x}+\sqrt{1-\sin x}\right)}

\displaystyle =\frac{(1+\sin x)+(1-\sin x)+2\sqrt{(1+\sin x)(1-\sin x)}} {(1+\sin x)-(1-\sin x)}

\displaystyle =\frac{2+2\sqrt{1-\sin^{2}x}}{2\sin x}

\displaystyle =\frac{1+\cos x}{\sin x}

\displaystyle =\frac{2\cos^{2}\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}}

\displaystyle =\cot\frac{x}{2}

\displaystyle \text{Therefore, equation }(i)\text{ becomes}

\displaystyle y=\cot^{-1}\!\left(\cot\frac{x}{2}\right)

\displaystyle y=\frac{x}{2} \qquad[\text{Since } \cot^{-1}(\cot\theta)=\theta,\ \theta\in(0,\pi)]

\displaystyle \text{Differentiating with respect to }x,

\displaystyle \frac{dy}{dx}=\frac{1}{2}

\displaystyle \text{Hence, } \frac{dy}{dx}\text{ is independent of }x.

\displaystyle \textbf{Question 39. }\ \text{If } y=\tan^{-1}\!\left(\frac{2x}{1-x^{2}}\right)+\sec^{-1}\!\left(\frac{1+x^{2}}{1-x^{2}}\right),\ x>0,\ \text{prove that }\frac{dy}{dx}=\frac{4}{1+x^{2}}
\displaystyle \text{Answer:}

\displaystyle \text{Here, } y=\tan^{-1}\!\left(\frac{2x}{1-x^{2}}\right) +\sec^{-1}\!\left(\frac{1+x^{2}}{1-x^{2}}\right)

\displaystyle y=\tan^{-1}\!\left(\frac{2x}{1-x^{2}}\right) +\cos^{-1}\!\left(\frac{1-x^{2}}{1+x^{2}}\right) \qquad[\text{Since } \sec^{-1}u=\cos^{-1}\!\left(\frac{1}{u}\right)]

\displaystyle \text{Put } x=\tan\theta

\displaystyle y=\tan^{-1}\!\left(\frac{2\tan\theta}{1-\tan^{2}\theta}\right) +\cos^{-1}\!\left(\frac{1-\tan^{2}\theta}{1+\tan^{2}\theta}\right)

\displaystyle y=\tan^{-1}(\tan2\theta)+\cos^{-1}(\cos2\theta)

\displaystyle \text{Here, } 0<x<1

\displaystyle \Rightarrow\ 0<\tan\theta<1

\displaystyle \Rightarrow\ 0<\theta<\frac{\pi}{4}

\displaystyle \Rightarrow\ 0<2\theta<\frac{\pi}{2}

\displaystyle y=2\theta+2\theta \quad[\text{Since } \tan^{-1}(\tan\alpha)=\alpha,\ \alpha\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right),\ \cos^{-1}(\cos\beta)=\beta,\ \beta\in[0,\pi]]

\displaystyle y=4\theta

\displaystyle \Rightarrow\ y=4\tan^{-1}x\qquad[\text{Since } x=\tan\theta]

\displaystyle \text{Differentiating with respect to }x,

\displaystyle \frac{dy}{dx} =4\cdot\frac{1}{1+x^{2}}

\displaystyle \frac{dy}{dx}=\frac{4}{1+x^{2}}

\displaystyle \textbf{Question 40. }\ \text{If } y=\sec^{-1}\!\left(\frac{x+1}{x-1}\right)+\sin^{-1}\!\left(\frac{x-1}{x+1}\right),\ x>0,\ \text{find }\frac{dy}{dx}
\displaystyle \text{Answer:}

\displaystyle \text{Here, } y=\sec^{-1}\!\left(\frac{x+1}{x-1}\right) +\sin^{-1}\!\left(\frac{x-1}{x+1}\right)

\displaystyle y=\cos^{-1}\!\left(\frac{x-1}{x+1}\right) +\sin^{-1}\!\left(\frac{x-1}{x+1}\right) \qquad[\text{Since } \sec^{-1}u=\cos^{-1}\!\left(\frac{1}{u}\right)]

\displaystyle y=\frac{\pi}{2} \qquad[\text{Since } \cos^{-1}\alpha+\sin^{-1}\alpha=\frac{\pi}{2}]

\displaystyle \text{Differentiating with respect to }x,

\displaystyle \frac{dy}{dx}=0

\displaystyle \textbf{Question 41. }\ \text{If } y=\sin\!\left[2\tan^{-1}\!\left(\sqrt{\frac{1-x}{1+x}}\right)\right],\ \text{find }\frac{dy}{dx}
\displaystyle \text{Answer:}

\displaystyle \text{Here, } y=\sin\!\left[2\tan^{-1}\!\left(\sqrt{\frac{1-x}{1+x}}\right)\right]

\displaystyle \text{Put } x=\cos2\theta

\displaystyle y=\sin\!\left[2\tan^{-1}\!\left(\sqrt{\frac{1-\cos2\theta}{1+\cos2\theta}}\right)\right]

\displaystyle y=\sin\!\left[2\tan^{-1}\!\left(\sqrt{\frac{2\sin^{2}\theta}{2\cos^{2}\theta}}\right)\right]

\displaystyle y=\sin\!\left[2\tan^{-1}(\tan\theta)\right]

\displaystyle y=\sin(2\theta) \qquad[\text{Since } \tan^{-1}(\tan\theta)=\theta,\ \theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)]

\displaystyle y=\sin\!\left(2\cdot\frac{1}{2}\cos^{-1}x\right) \qquad[\text{Since } x=\cos2\theta]

\displaystyle y=\sin(\cos^{-1}x)

\displaystyle y=\sqrt{1-x^{2}}

\displaystyle \text{Differentiating with respect to }x\text{ using chain rule,}

\displaystyle \frac{dy}{dx} =\frac{1}{2\sqrt{1-x^{2}}}\cdot\frac{d}{dx}(1-x^{2})

\displaystyle \frac{dy}{dx} =\frac{1}{2\sqrt{1-x^{2}}}\cdot(-2x)

\displaystyle \frac{dy}{dx}=-\frac{x}{\sqrt{1-x^{2}}}

\displaystyle \textbf{Question 42. }\ \text{If } y=\cos^{-1}(2x)+2\cos^{-1}\!\left(\sqrt{1-4x^{2}}\right),\ 0<x<\frac{1}{2},\ \text{find }\frac{dy}{dx}
\displaystyle \text{Answer:}

\displaystyle \text{Here, } y=\cos^{-1}(2x)+2\cos^{-1}\!\left(\sqrt{1-4x^{2}}\right)

\displaystyle \text{Put } 2x=\cos\theta

\displaystyle y=\cos^{-1}(\cos\theta)+2\cos^{-1}\!\left(\sqrt{1-\cos^{2}\theta}\right)

\displaystyle y=\cos^{-1}(\cos\theta)+2\cos^{-1}(\sin\theta)

\displaystyle y=\cos^{-1}(\cos\theta)+2\cos^{-1}\!\left(\cos\!\left(\frac{\pi}{2}-\theta\right)\right)\qquad\ldots(i)

\displaystyle \text{Here, } 0<x<\frac{1}{2}

\displaystyle \Rightarrow\ 0<2x<1

\displaystyle \Rightarrow\ 0<\cos\theta<1

\displaystyle \Rightarrow\ 0<\theta<\frac{\pi}{2}

\displaystyle \Rightarrow\ 0<\frac{\pi}{2}-\theta<\frac{\pi}{2}

\displaystyle \text{So, from equation }(i)

\displaystyle y=\theta+2\!\left(\frac{\pi}{2}-\theta\right) \quad[\text{Since } \cos^{-1}(\cos\alpha)=\alpha,\ \alpha\in[0,\pi]]

\displaystyle y=\pi-\theta

\displaystyle y=\pi-\cos^{-1}(2x)\qquad[\text{Since } 2x=\cos\theta]

\displaystyle \text{Differentiating with respect to }x\text{ using chain rule,}

\displaystyle \frac{dy}{dx} =0-\left(-\frac{1}{\sqrt{1-(2x)^{2}}}\right)\frac{d}{dx}(2x)

\displaystyle \frac{dy}{dx} =\frac{2}{\sqrt{1-4x^{2}}}

\displaystyle \textbf{Question 43. }\ \text{If the derivative of }\tan^{-1}(a+bx)\ \text{takes the value }1\ \text{at }x=0,\ \text{prove that }1+a^{2}=b
\displaystyle \text{Answer:}

\displaystyle \text{Here, } \left.\frac{d}{dx}\!\left[\tan^{-1}(a+bx)\right]\right|_{x=0}=1

\displaystyle \Rightarrow \left[\frac{1}{1+(a+bx)^2}\cdot\frac{d}{dx}(a+bx)\right]_{x=0}=1

\displaystyle \Rightarrow \left[\frac{1}{1+(a+bx)^2}\cdot b\right]_{x=0}=1

\displaystyle \Rightarrow \frac{b}{1+(a+0)^2}=1

\displaystyle \Rightarrow \frac{b}{1+a^2}=1

\displaystyle \Rightarrow b=1+a^2

\displaystyle \textbf{Question 44. }\ \text{If } y=\cos^{-1}(2x)+2\cos^{-1}\!\left(\sqrt{1-4x^{2}}\right),\ -\frac{1}{2}<x<0,\ \text{find }\frac{dy}{dx}
\displaystyle \text{Answer:}

\displaystyle \text{Here, } y=\cos^{-1}(2x)+2\cos^{-1}\!\left(\sqrt{1-4x^{2}}\right)

\displaystyle \text{Put } 2x=\cos\theta

\displaystyle y=\cos^{-1}(\cos\theta)+2\cos^{-1}\!\left(\sqrt{1-\cos^{2}\theta}\right)

\displaystyle y=\cos^{-1}(\cos\theta)+2\cos^{-1}(\sin\theta)

\displaystyle y=\cos^{-1}(\cos\theta)+2\cos^{-1}\!\left(\cos\!\left(\frac{\pi}{2}-\theta\right)\right)\qquad\ldots(1)

\displaystyle \text{Now, } -\frac{1}{2}<x<0

\displaystyle \Rightarrow -1<2x<0

\displaystyle \Rightarrow -1<\cos\theta<0

\displaystyle \Rightarrow \frac{\pi}{2}<\theta<\pi

\displaystyle \Rightarrow -\frac{\pi}{2}<\frac{\pi}{2}-\theta<0

\displaystyle \text{So, from equation }(1)

\displaystyle y=\theta+2\!\left[-\!\left(\frac{\pi}{2}-\theta\right)\right] \quad[\text{Since } \cos^{-1}(\cos\alpha)=\alpha\ \text{for }\alpha\in[0,\pi],\ \cos^{-1}(\cos\beta)=-\beta\ \text{for }\beta\in[-\pi,0]]

\displaystyle y=\theta-\pi+2\theta

\displaystyle y=-\pi+3\theta

\displaystyle y=-\pi+3\cos^{-1}(2x)\qquad[\text{Since } 2x=\cos\theta]

\displaystyle \text{Differentiating with respect to }x\text{ using chain rule,}

\displaystyle \frac{dy}{dx} =0+3\!\left(-\frac{1}{\sqrt{1-(2x)^{2}}}\right)\frac{d}{dx}(2x)

\displaystyle \frac{dy}{dx} =-\frac{6}{\sqrt{1-4x^{2}}}

\displaystyle \textbf{Question 45. }\ \text{If } y=\tan^{-1}\!\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right),\ \text{find }\frac{dy}{dx}
\displaystyle \text{Answer:}

\displaystyle \text{Here, } y=\tan^{-1}\!\left( \frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}} \right)

\displaystyle \text{Put } x=\cos2\theta

\displaystyle y=\tan^{-1}\!\left( \frac{\sqrt{1+\cos2\theta}-\sqrt{1-\cos2\theta}} {\sqrt{1+\cos2\theta}+\sqrt{1-\cos2\theta}} \right)

\displaystyle y=\tan^{-1}\!\left( \frac{\sqrt{2\cos^{2}\theta}-\sqrt{2\sin^{2}\theta}} {\sqrt{2\cos^{2}\theta}+\sqrt{2\sin^{2}\theta}} \right)

\displaystyle y=\tan^{-1}\!\left( \frac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta} \right)

\displaystyle y=\tan^{-1}\!\left( \frac{1-\tan\theta}{1+\tan\theta} \right)

\displaystyle y=\tan^{-1}\!\left( \frac{\tan\frac{\pi}{4}-\tan\theta} {1+\tan\frac{\pi}{4}\tan\theta} \right)

\displaystyle y=\tan^{-1}\!\left[\tan\!\left(\frac{\pi}{4}-\theta\right)\right]

\displaystyle y=\frac{\pi}{4}-\theta \qquad\left[\text{Since } \theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\right]

\displaystyle y=\frac{\pi}{4}-\frac{1}{2}\cos^{-1}x \qquad[\text{Since } x=\cos2\theta]

\displaystyle \text{Differentiating with respect to }x,

\displaystyle \frac{dy}{dx} =0-\frac{1}{2}\!\left(-\frac{1}{\sqrt{1-x^{2}}}\right)

\displaystyle \frac{dy}{dx}=\frac{1}{2\sqrt{1-x^{2}}}

\displaystyle \textbf{Question 46. }\ \text{If } y=\cos^{-1}\!\left(\frac{2x-3\sqrt{1-x^{2}}}{\sqrt{13}}\right),\ \text{find }\frac{dy}{dx}
\displaystyle \text{Answer:}

\displaystyle \text{Let } y=\cos^{-1}\!\left(\frac{2x-3\sqrt{1-x^{2}}}{\sqrt{13}}\right)

\displaystyle \text{Put } x=\cos\theta

\displaystyle y=\cos^{-1}\!\left(\frac{2\cos\theta-3\sqrt{1-\cos^{2}\theta}}{\sqrt{13}}\right)

\displaystyle y=\cos^{-1}\!\left(\frac{2\cos\theta-3\sin\theta}{\sqrt{13}}\right)

\displaystyle y=\cos^{-1}\!\left(\cos\theta\frac{2}{\sqrt{13}}-\sin\theta\frac{3}{\sqrt{13}}\right)

\displaystyle \text{Let } \cos\phi=\frac{2}{\sqrt{13}},\quad \sin\phi=\frac{3}{\sqrt{13}}

\displaystyle y=\cos^{-1}\!\left(\cos\theta\cos\phi-\sin\theta\sin\phi\right)

\displaystyle y=\cos^{-1}\!\left(\cos(\theta+\phi)\right)\qquad\ldots(i)

\displaystyle y=\theta+\phi \quad[\text{Since } \cos^{-1}(\cos\alpha)=\alpha,\ \alpha\in[0,\pi]]

\displaystyle y=\cos^{-1}x+\phi

\displaystyle \text{Differentiating with respect to }x,

\displaystyle \frac{dy}{dx} =\frac{d}{dx}(\cos^{-1}x)+0

\displaystyle \frac{dy}{dx} =-\frac{1}{\sqrt{1-x^{2}}}

\displaystyle \textbf{Question 47. }\ \text{Differentiate }\sin^{-1}\!\left(\frac{2^{x+1}-3^{x}}{1+36^{x}}\right)\ \text{with respect to }x
\displaystyle \text{Answer:}

\displaystyle \text{We have, } y=\sin^{-1}\!\left(\frac{2^{x+1}\cdot 3^{x}}{1+36^{x}}\right)

\displaystyle y=\sin^{-1}\!\left(\frac{2\cdot 6^{x}}{1+6^{2x}}\right)

\displaystyle \text{Put } 6^{x}=\tan\theta

\displaystyle \Rightarrow \theta=\tan^{-1}(6^{x})

\displaystyle y=\sin^{-1}\!\left(\frac{2\tan\theta}{1+\tan^{2}\theta}\right)

\displaystyle y=\sin^{-1}(\sin 2\theta)

\displaystyle y=2\theta \qquad[\text{Since } \sin^{-1}(\sin\alpha)=\alpha,\ \alpha\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]]

\displaystyle y=2\tan^{-1}(6^{x})

\displaystyle \text{Differentiating with respect to }x,

\displaystyle \frac{dy}{dx} =2\cdot\frac{1}{1+(6^{x})^{2}}\cdot\frac{d}{dx}(6^{x})

\displaystyle \frac{dy}{dx} =\frac{2}{1+6^{2x}}\cdot 6^{x}\ln 6

\displaystyle \frac{dy}{dx} =\frac{2\cdot 6^{x}\ln 6}{1+36^{x}}

\displaystyle \textbf{Question 48. }\ \text{If } y=\sin^{-1}\!\left(6x\sqrt{1-9x^{2}}\right),\ -\frac{1}{3\sqrt{2}}<x<\frac{1}{3\sqrt{2}},\ \text{find }\frac{dy}{dx}
\displaystyle \text{Answer:}

\displaystyle \text{We have } y=\sin^{-1}\!\left(6x\sqrt{1-9x^{2}}\right), \qquad -\frac{1}{3\sqrt{2}}<x<\frac{1}{3\sqrt{2}}

\displaystyle \text{So, } \frac{dy}{dx} =\frac{1}{\sqrt{1-(6x\sqrt{1-9x^{2}})^{2}}} \cdot \frac{d}{dx}\!\left(6x\sqrt{1-9x^{2}}\right)

\displaystyle =\frac{1}{\sqrt{1-36x^{2}(1-9x^{2})}} \cdot \frac{d}{dx}\!\left(6x\sqrt{1-9x^{2}}\right)

\displaystyle =\frac{1}{\sqrt{(1-18x^{2})^{2}}} \cdot \frac{d}{dx}\!\left(6x\sqrt{1-9x^{2}}\right)

\displaystyle =\frac{1}{1-18x^{2}} \cdot \left[ 6\sqrt{1-9x^{2}}+ 6x\cdot\frac{-18x}{2\sqrt{1-9x^{2}}} \right]

\displaystyle =\frac{1}{1-18x^{2}} \cdot \frac{6(1-18x^{2})}{\sqrt{1-9x^{2}}}

\displaystyle \frac{dy}{dx} =\frac{6}{\sqrt{1-9x^{2}}}

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