\displaystyle \text{Find } \frac{dy}{dx}, \text{ when } 

\displaystyle \textbf{Question 1: }~x=at^2\text{ and }y=2at
\displaystyle \text{Answer:}

\displaystyle \text{We have, } x = at^{2} \text{ and } y = 2at

\displaystyle \Rightarrow \frac{dx}{dt} = 2at \text{ and } \frac{dy}{dt} = 2a

\displaystyle \therefore \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2a}{2at} = \frac{1}{t}

\displaystyle \textbf{Question 2: }~x=a(\theta+\sin\theta)\text{ and }y=a(1-\cos\theta)
\displaystyle \text{Answer:}

\displaystyle \text{We have, } x = a(\theta + \sin\theta) \text{ and } y = a(1 - \cos\theta)

\displaystyle \Rightarrow \frac{dx}{d\theta} = a(1 + \cos\theta) \text{ and } \frac{dy}{d\theta} = a\sin\theta

\displaystyle \therefore \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{a\sin\theta}{a(1 + \cos\theta)} = \frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\cos^{2}\frac{\theta}{2}} = \tan\frac{\theta}{2}

\displaystyle \textbf{Question 3: }~x=a\cos\theta\text{ and }y=b\sin\theta
\displaystyle \text{Answer:}

\displaystyle \text{We have, } x = a\cos\theta \text{ and } y = b\sin\theta

\displaystyle \Rightarrow \frac{dx}{d\theta} = -a\sin\theta \text{ and } \frac{dy}{d\theta} = b\cos\theta

\displaystyle \therefore \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{b\cos\theta}{-a\sin\theta} = -\frac{b}{a}\cot\theta

\displaystyle \textbf{Question 4: }~x=ae^{\theta}(\sin\theta-\cos\theta),\;y=ae^{\theta}(\sin\theta+\cos\theta)
\displaystyle \text{Answer:}

\displaystyle \text{We have, } x = ae^{\theta}(\sin\theta - \cos\theta) \text{ and } y = ae^{\theta}(\sin\theta + \cos\theta)

\displaystyle \Rightarrow \frac{dx}{d\theta} = a\left[e^{\theta}\frac{d}{d\theta}(\sin\theta - \cos\theta) + (\sin\theta - \cos\theta)\frac{d}{d\theta}(e^{\theta})\right] \text{ and } \frac{dy}{d\theta} = a\left[e^{\theta}\frac{d}{d\theta}(\sin\theta + \cos\theta) + (\sin\theta + \cos\theta)\frac{d}{d\theta}(e^{\theta})\right]

\displaystyle \Rightarrow \frac{dx}{d\theta} = a\left[e^{\theta}(\cos\theta + \sin\theta) + (\sin\theta - \cos\theta)e^{\theta}\right] \text{ and } \frac{dy}{d\theta} = a\left[e^{\theta}(\cos\theta - \sin\theta) + (\sin\theta + \cos\theta)e^{\theta}\right]

\displaystyle \Rightarrow \frac{dx}{d\theta} = a\left[2e^{\theta}\sin\theta\right] \text{ and } \frac{dy}{d\theta} = a\left[2e^{\theta}\cos\theta\right]

\displaystyle \therefore \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{a(2e^{\theta}\cos\theta)}{a(2e^{\theta}\sin\theta)} = \cot\theta

\displaystyle \textbf{Question 5: }~x=b\sin^2\theta\text{ and }y=a\cos^2\theta
\displaystyle \text{Answer:}

\displaystyle \text{We have, } x = b\sin^{2}\theta \text{ and } y = a\cos^{2}\theta

\displaystyle \therefore \frac{dx}{d\theta} = \frac{d}{d\theta}(b\sin^{2}\theta) = 2b\sin\theta\cos\theta

\displaystyle \text{and,}

\displaystyle \frac{dy}{d\theta} = \frac{d}{d\theta}(a\cos^{2}\theta) = -2a\cos\theta\sin\theta

\displaystyle \therefore \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{-2a\cos\theta\sin\theta}{2b\sin\theta\cos\theta} = -\frac{a}{b}

\displaystyle \textbf{Question 6: }~x=a(1-\cos\theta)\text{ and }y=a(\theta+\sin\theta)\text{ at }\theta=\frac{\pi}{2}
\displaystyle \text{Answer:}

\displaystyle \text{We have, } x = a(1 - \cos\theta) \text{ and } y = a(\theta + \sin\theta)

\displaystyle \therefore \frac{dx}{d\theta} = \frac{d}{d\theta}[a(1 - \cos\theta)] = a(\sin\theta)

\displaystyle \text{and}

\displaystyle \frac{dy}{d\theta} = \frac{d}{d\theta}[a(\theta + \sin\theta)] = a(1 + \cos\theta)

\displaystyle \therefore \left[\frac{dy}{dx}\right]_{\theta=\frac{\pi}{2}} = \left[\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}\right]_{\theta=\frac{\pi}{2}} = \left[\frac{a(1+\cos\theta)}{a(\sin\theta)}\right]_{\theta=\frac{\pi}{2}} = \frac{a(1+0)}{a} = 1

\displaystyle \textbf{Question 7: }~x=\frac{e^{t}+e^{-t}}{2}\text{ and }y=\frac{e^{t}-e^{-t}}{2}
\displaystyle \text{Answer:}

\displaystyle \text{We have, } x = \frac{e^{t} + e^{-t}}{2} \text{ and } y = \frac{e^{t} - e^{-t}}{2}

\displaystyle \Rightarrow \frac{dx}{dt} = \frac{1}{2}\left[\frac{d}{dt}(e^{t}) + \frac{d}{dt}(e^{-t})\right] \text{ and } \frac{dy}{dt} = \frac{1}{2}\left[\frac{d}{dt}(e^{t}) - \frac{d}{dt}(e^{-t})\right]

\displaystyle \Rightarrow \frac{dx}{dt} = \frac{1}{2}\left[e^{t} + e^{-t}\frac{d}{dt}(-t)\right] \text{ and } \frac{dy}{dt} = \frac{1}{2}\left[e^{t} - e^{-t}\frac{d}{dt}(-t)\right]

\displaystyle \Rightarrow \frac{dx}{dt} = \frac{1}{2}(e^{t} - e^{-t}) = y \text{ and } \frac{dy}{dt} = \frac{1}{2}(e^{t} + e^{-t}) = x

\displaystyle \therefore \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{x}{y}

\displaystyle \textbf{Question 8: }~x=\frac{3at}{1+t^2}\text{ and }y=\frac{3at^2}{1+t^2}
\displaystyle \text{Answer:}

\displaystyle \text{We have, } x = \frac{3at}{1+t^{2}}

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

\displaystyle \frac{dx}{dt} =\left[\frac{(1+t^{2})\frac{d}{dt}(3at)-3at\frac{d}{dt}(1+t^{2})}{(1+t^{2})^{2}}\right] \ \ldots\ldots\ldots\ldots\ \text{[using quotient rule]}

\displaystyle \Rightarrow \frac{dx}{dt} =\left[\frac{(1+t^{2})(3a)-3at(2t)}{(1+t^{2})^{2}}\right]

\displaystyle \Rightarrow \frac{dx}{dt} =\left[\frac{3a+3at^{2}-6at^{2}}{(1+t^{2})^{2}}\right]

\displaystyle \Rightarrow \frac{dx}{dt} =\left[\frac{3a-3at^{2}}{(1-t^{2})^{2}}\right]

\displaystyle \Rightarrow \frac{dx}{dt} =\frac{3a(1-t^{2})}{(1+t^{2})^{2}} \ \ldots\ (i)

\displaystyle \text{and, } y = \frac{3at^{2}}{1+t^{2}}

\displaystyle \text{Differentiating it with respect to } t,

\displaystyle \frac{dx}{dt} =\left[\frac{(1+t^{2})\frac{d}{dt}(3at^{2})-3at^{2}\frac{d}{dt}(1+t^{2})}{(1+t^{2})^{2}}\right] \ \ldots\ldots\ldots\ldots\ \text{[using quotient rule]}

\displaystyle \Rightarrow \frac{dx}{dt} =\left[\frac{(1+t^{2})(6at)-3at^{2}(2t)}{(1+t^{2})^{2}}\right]

\displaystyle \Rightarrow \frac{dx}{dt} =\left[\frac{6at+6at^{3}-6at^{3}}{(1+t^{2})^{2}}\right]

\displaystyle \Rightarrow \frac{dx}{dt} =\frac{6at}{(1+t^{2})^{2}} \ \ldots\ (ii)

\displaystyle \text{Dividing equation (ii) by (i),}

\displaystyle \frac{\frac{dy}{dt}}{\frac{dx}{dt}} =\frac{6at}{(1+t^{2})^{2}} \times \frac{(1+t^{2})^{2}}{3a(1-t^{2})} =\frac{2t}{1-t^{2}}

\displaystyle \textbf{Question 9: }~x=a(\cos\theta+\theta\sin\theta)\text{ and }y=a(\sin\theta-\theta\cos\theta)
\displaystyle \text{Answer:}

\displaystyle \text{We have, } x = a(\cos\theta + \theta\sin\theta) \text{ and } y = a(\sin\theta - \theta\cos\theta)

\displaystyle \Rightarrow \frac{dx}{d\theta} = a\left[\frac{d}{d\theta}\cos\theta + \frac{d}{d\theta}(\theta\sin\theta)\right] \text{ and } \frac{dy}{d\theta} = a\left[\frac{d}{d\theta}(\sin\theta) - \frac{d}{d\theta}(\theta\cos\theta)\right]

\displaystyle \Rightarrow \frac{dx}{d\theta} = a\left[-\sin\theta + \theta\frac{d}{d\theta}(\sin\theta) + \sin\theta\frac{d}{d\theta}(\theta)\right] \text{ and } \frac{dy}{d\theta} = a\left[\cos\theta - \left\{\theta\frac{d}{d\theta}(\cos\theta) + \cos\theta\frac{d}{d\theta}(\theta)\right\}\right]

\displaystyle \Rightarrow \frac{dx}{d\theta} = a[-\sin\theta + \theta\cos\theta] \text{ and } \frac{dy}{d\theta} = a[\cos\theta + \theta\sin\theta - \cos\theta]

\displaystyle \Rightarrow \frac{dx}{d\theta} = a\theta\cos\theta \text{ and } \frac{dy}{d\theta} = a\theta\sin\theta

\displaystyle \therefore \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{a\theta\sin\theta}{a\theta\cos\theta} = \tan\theta

\displaystyle \textbf{Question 10: }~x=e^{\theta}\!\left(\theta+\frac{1}{\theta}\right)\text{ and }y=e^{-\theta}\!\left(\theta-\frac{1}{\theta}\right)
\displaystyle \text{Answer:}

\displaystyle \text{We have, } x = e^{\theta}\left(\theta + \frac{1}{\theta}\right)

\displaystyle \text{Differentiating it with respect to } \theta

\displaystyle \frac{dx}{d\theta} = e^{\theta}\frac{d}{d\theta}\left(\theta + \frac{1}{\theta}\right) + \left(\theta + \frac{1}{\theta}\right)\frac{d}{d\theta}\left(e^{\theta}\right) \ \text{[using product rule]}

\displaystyle \Rightarrow \frac{dx}{d\theta} = e^{\theta}\left(1-\frac{1}{\theta^{2}}\right) + \left(\frac{\theta^{2}+1}{\theta}\right)e^{\theta}

\displaystyle \Rightarrow \frac{dx}{d\theta} = e^{\theta}\left(1-\frac{1}{\theta^{2}}+\frac{\theta^{2}+1}{\theta}\right)

\displaystyle \Rightarrow \frac{dx}{d\theta} = e^{\theta}\left(\frac{\theta^{2}-1+\theta^{3}+\theta}{\theta^{2}}\right)

\displaystyle \Rightarrow \frac{dx}{d\theta} = \frac{e^{\theta}(\theta^{3}+\theta^{2}+\theta-1)}{\theta^{2}} \ \ldots\ (i)

\displaystyle \text{and,}

\displaystyle y = e^{-\theta}\left(\theta-\frac{1}{\theta}\right)

\displaystyle \text{Differentiating it with respect to } \theta \text{ using chain rule}

\displaystyle \frac{dy}{d\theta} = e^{-\theta}\frac{d}{d\theta}\left(\theta-\frac{1}{\theta}\right) + \left(\theta-\frac{1}{\theta}\right)\frac{d}{d\theta}\left(e^{-\theta}\right) \ \text{[using product rule]}

\displaystyle \Rightarrow \frac{dy}{d\theta} = e^{-\theta}\left(1+\frac{1}{\theta^{2}}\right) + \left(\theta-\frac{1}{\theta}\right)e^{-\theta}\frac{d}{d\theta}(-\theta)

\displaystyle \Rightarrow \frac{dy}{d\theta} = e^{-\theta}\left(1+\frac{1}{\theta^{2}}\right) + \left(\theta-\frac{1}{\theta}\right)e^{-\theta}(-1)

\displaystyle \Rightarrow \frac{dy}{d\theta} = e^{-\theta}\left(1+\frac{1}{\theta^{2}}-\theta+\frac{1}{\theta}\right)

\displaystyle \Rightarrow \frac{dy}{d\theta} = e^{-\theta}\left(\frac{\theta^{2}+1-\theta^{3}+\theta}{\theta^{2}}\right)

\displaystyle \Rightarrow \frac{dy}{d\theta} = \frac{e^{-\theta}(-\theta^{3}+\theta^{2}+\theta+1)}{\theta^{2}} \ \ldots\ (ii)

\displaystyle \text{Dividing equation (ii) by (i),}

\displaystyle \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{e^{-\theta}\left(\frac{\theta^{2}-\theta^{3}+\theta+1}{\theta^{2}}\right)} {\frac{e^{\theta}(\theta^{3}+\theta^{2}+\theta-1)}{\theta^{2}}}

\displaystyle = e^{-2\theta}\left(\frac{\theta^{2}-\theta^{3}+\theta+1}{\theta^{3}+\theta^{2}+\theta-1}\right)

\displaystyle \textbf{Question 11: }~x=\frac{2t}{1+t^2}\text{ and }y=\frac{1-t^2}{1+t^2}
\displaystyle \text{Answer:}

\displaystyle \text{We have, } x=\frac{2t}{1+t^{2}}

\displaystyle \Rightarrow \frac{dx}{dt} =\left[\frac{(1+t^{2})\frac{d}{dt}(2t)-2t\frac{d}{dt}(1+t^{2})}{(1+t^{2})^{2}}\right] \ \text{[using quotient rule]}

\displaystyle \Rightarrow \frac{dx}{dt} =\left[\frac{(1+t^{2})(2)-2t(2t)}{(1+t^{2})^{2}}\right]

\displaystyle \Rightarrow \frac{dx}{dt} =\left[\frac{2+2t^{2}-4t^{2}}{(1+t^{2})^{2}}\right]

\displaystyle \Rightarrow \frac{dx}{dt} =\left[\frac{2-2t^{2}}{(1+t^{2})^{2}}\right] \ \ldots (i)

\displaystyle \text{and,}

\displaystyle y=\frac{1-t^{2}}{1+t^{2}}

\displaystyle \Rightarrow \frac{dy}{dt} =\left[\frac{(1+t^{2})\frac{d}{dt}(1-t^{2})-(1-t^{2})\frac{d}{dt}(1+t^{2})}{(1+t^{2})^{2}}\right]

\displaystyle \Rightarrow \frac{dy}{dt} =\left[\frac{(1+t^{2})(-2t)-(1-t^{2})(2t)}{(1+t^{2})^{2}}\right]

\displaystyle \Rightarrow \frac{dy}{dt} =\left[\frac{-4t}{(1+t^{2})^{2}}\right] \ \ldots (ii)

\displaystyle \text{Dividing equation (ii) by (i), we get,}

\displaystyle \frac{\frac{dy}{dt}}{\frac{dx}{dt}} =\frac{-4t}{(1+t^{2})^{2}} \times\frac{(1+t^{2})^{2}}{2(1-t^{2})}

\displaystyle \Rightarrow \frac{dy}{dx} =\frac{-2t}{1-t^{2}}

\displaystyle \Rightarrow \frac{dy}{dx} =-\frac{x}{y} \quad \left[\because \frac{x}{y} =\frac{2t}{1+t^{2}}\times\frac{1+t^{2}}{1-t^{2}} =\frac{2t}{1-t^{2}}\right]

\displaystyle \textbf{Question 12: }~x=\cos^{-1}\!\left(\frac{1}{\sqrt{1+t^2}}\right)\text{ and }y=\sin^{-1}\!\left(\frac{t}{\sqrt{1+t^2}}\right),\;t\in\mathbb{R}
\displaystyle \text{Answer:}

\displaystyle \text{We have, } x = \cos^{-1}\left(\frac{1}{\sqrt{1+t^{2}}}\right)

\displaystyle \Rightarrow \frac{dx}{dt} = \frac{-1}{\sqrt{1-\left(\frac{1}{\sqrt{1+t^{2}}}\right)^{2}}}\; \frac{d}{dt}\left(\frac{1}{\sqrt{1+t^{2}}}\right)

\displaystyle \Rightarrow \frac{dx}{dt} = \frac{-1}{\sqrt{1-\frac{1}{(1+t^{2})}}}\left\{ \frac{-1}{2(1+t^{2})^{\frac{3}{2}}}\frac{d}{dt}(1+t^{2})\right\}

\displaystyle \Rightarrow \frac{dx}{dt} = \frac{(1+t^{2})^{\frac{1}{2}}}{\sqrt{1+t^{2}-1}}\times \frac{1}{2(1+t^{2})^{\frac{3}{2}}}(2t)

\displaystyle \Rightarrow \frac{dx}{dt} = \frac{t}{\sqrt{t^{2}}\times(1+t^{2})}

\displaystyle \Rightarrow \frac{dx}{dt} = \frac{1}{1+t^{2}} \ \ldots\ (i)

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

\displaystyle \Rightarrow \frac{dy}{dt} = \frac{1}{\sqrt{1-\left(\frac{1}{\sqrt{1+t^{2}}}\right)^{2}}}\; \frac{d}{dt}\left(\frac{1}{\sqrt{1+t^{2}}}\right)

\displaystyle \Rightarrow \frac{dy}{dt} = \frac{1}{\sqrt{1-\frac{1}{(1+t^{2})}}}\left\{ \frac{-1}{2(1+t^{2})^{\frac{3}{2}}}\frac{d}{dt}(1+t^{2})\right\}

\displaystyle \Rightarrow \frac{dx}{dt} = \frac{(1+t^{2})^{\frac{1}{2}}}{\sqrt{1+t^{2}-1}}\times \frac{-1}{2(1+t^{2})^{\frac{3}{2}}}(2t)

\displaystyle \Rightarrow \frac{dx}{dt} = \frac{-1}{2\sqrt{t^{2}}\times(1+t^{2})}(2t)

\displaystyle \Rightarrow \frac{dx}{dt} = \frac{-1}{1+t^{2}} \ \ldots\ (ii)

\displaystyle \text{Dividing equation (ii) by (i),}

\displaystyle \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-1}{(1+t^{2})}\times \frac{(1+t^{2})}{1}

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

\displaystyle \textbf{Question 13: }~x=\frac{1-t^2}{1+t^2}\text{ and }y=\frac{2t}{1+t^2}
\displaystyle \text{Answer:}

\displaystyle \text{We have, } y = \frac{2t}{1+t^{2}}

\displaystyle \Rightarrow \frac{dy}{dt} =\left[\frac{(1+t^{2})\frac{d}{dt}(2t)-2t\frac{d}{dt}(1+t^{2})}{(1+t^{2})^{2}}\right] \ \text{[using quotient rule]}

\displaystyle \Rightarrow \frac{dy}{dt} =\left[\frac{(1+t^{2})(2)-2t(2t)}{(1+t^{2})^{2}}\right]

\displaystyle \Rightarrow \frac{dy}{dt} =\left[\frac{2+2t^{2}-4t^{2}}{(1+t^{2})^{2}}\right]

\displaystyle \Rightarrow \frac{dy}{dt} =\left[\frac{2-2t^{2}}{(1+t^{2})^{2}}\right]\ \ldots\ (i)

\displaystyle \text{and,}

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

\displaystyle \Rightarrow \frac{dx}{dt} =\left[\frac{(1+t^{2})\frac{d}{dt}(1-t^{2})-(1-t^{2})\frac{d}{dt}(1+t^{2})}{(1+t^{2})^{2}}\right]

\displaystyle \Rightarrow \frac{dx}{dt} =\left[\frac{(1+t^{2})(-2t)-(1-t^{2})(2t)}{(1+t^{2})^{2}}\right]

\displaystyle \Rightarrow \frac{dx}{dt} =\left[\frac{-4t}{(1+t^{2})^{2}}\right]\ \ldots\ (ii)

\displaystyle \text{Dividing equation (i) by (ii), we get ,}

\displaystyle \frac{\frac{dy}{dt}}{\frac{dx}{dt}} =\frac{2(1-t^{2})}{(1+t^{2})^{2}} \times \frac{(1+t^{2})^{2}}{-4t}

\displaystyle \Rightarrow \frac{dy}{dx} =\frac{2(1-t^{2})}{-4t}

\displaystyle \Rightarrow \frac{dy}{dx} =\frac{t^{2}-1}{2t}

\displaystyle \textbf{Question 14: }~\text{If }x=2\cos\theta-\cos2\theta\text{ and }y=2\sin\theta-\sin2\theta,\text{ prove that }\frac{dy}{dx}=\tan\!\left(\frac{3\theta}{2}\right)
\displaystyle \text{Answer:}

\displaystyle \text{We have, } x = 2\cos\theta - \cos 2\theta

\displaystyle \Rightarrow \frac{dx}{d\theta} = 2(-\sin\theta) - \frac{d}{d\theta}(\cos 2\theta)

\displaystyle \Rightarrow \frac{dx}{d\theta} = -2\sin\theta + 2\sin 2\theta

\displaystyle \Rightarrow \frac{dx}{d\theta} = 2(\sin 2\theta - \sin\theta) \quad \text{...(i)}

\displaystyle \text{and,}

\displaystyle y = 2\sin\theta - \sin 2\theta

\displaystyle \Rightarrow \frac{dy}{d\theta} = 2\cos\theta - \frac{d}{d\theta}(\sin 2\theta)

\displaystyle \Rightarrow \frac{dy}{d\theta} = 2\cos\theta - 2\cos 2\theta

\displaystyle \Rightarrow \frac{dy}{d\theta} = 2(\cos\theta - \cos 2\theta) \quad \text{...(ii)}

\displaystyle \text{Dividing equation (ii) by equation (i),}

\displaystyle \frac{dy}{dx} = \frac{2(\cos\theta - \cos 2\theta)} {2(\sin 2\theta - \sin\theta)}

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{\cos\theta - \cos 2\theta}{\sin 2\theta - \sin\theta}

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{-2\sin\left(\frac{3\theta}{2}\right)\sin\left(-\frac{\theta}{2}\right)} {2\cos\left(\frac{3\theta}{2}\right)\sin\left(\frac{\theta}{2}\right)}

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{\sin\left(\frac{3\theta}{2}\right)} {\cos\left(\frac{3\theta}{2}\right)}

\displaystyle \therefore \frac{dy}{dx} = \tan\left(\frac{3\theta}{2}\right)

\displaystyle \textbf{Question 15: }~\text{If }x=e^{\cos2t}\text{ and }y=e^{\sin2t},\text{ prove that }\frac{dy}{dx}=-\frac{y\log x}{x\log y}\;
\displaystyle \text{Answer:}

\displaystyle \text{We have, } x = e^{\cos 2t} \text{ and } y = e^{\sin 2t}

\displaystyle \Rightarrow \frac{dx}{dt} = \frac{d}{dt}\left(e^{\cos 2t}\right) \text{ and } \frac{dy}{dt} = \frac{d}{dt}\left(e^{\sin 2t}\right)

\displaystyle \Rightarrow \frac{dx}{dt} = e^{\cos 2t}\frac{d}{dt}(\cos 2t) \text{ and } \frac{dy}{dt} = e^{\sin 2t}\frac{d}{dt}(\sin 2t)

\displaystyle \Rightarrow \frac{dx}{dt} = e^{\cos 2t}(-\sin 2t)\frac{d}{dt}(2t) \text{ and } \frac{dy}{dt} = e^{\sin 2t}(\cos 2t)\frac{d}{dt}(2t)

\displaystyle \Rightarrow \frac{dx}{dt} = -2\sin 2t\,e^{\cos 2t} \text{ and } \frac{dy}{dt} = 2\cos 2t\,e^{\sin 2t}

\displaystyle \therefore \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2\cos 2t\,e^{\sin 2t}}{-2\sin 2t\,e^{\cos 2t}}

\displaystyle \Rightarrow \frac{dy}{dx} = -\frac{e^{\sin 2t}}{e^{\cos 2t}}\cdot\frac{\cos 2t}{\sin 2t}

\displaystyle \Rightarrow \frac{dy}{dx} = -\frac{y\log x}{x\log y} \quad \left[\because\ x=e^{\cos 2t}\Rightarrow\log x=\cos 2t,\; y=e^{\sin 2t}\Rightarrow\log y=\sin 2t\right]

\displaystyle \textbf{Question 16: }~\text{If }x=\cos t\text{ and }y=\sin t,\text{ prove that }\frac{dy}{dx}=\frac{1}{\sqrt{3}}\text{ at }t=\frac{2\pi}{3}
\displaystyle \text{Answer:}

\displaystyle \text{We have, } x = \cos t \text{ and } y = \sin t

\displaystyle \Rightarrow \frac{dx}{dt} = \frac{d}{dt}(\cos t) \text{ and } \frac{dy}{dt} = \frac{d}{dt}(\sin t)

\displaystyle \Rightarrow \frac{dx}{dt} = -\sin t \text{ and } \frac{dy}{dt} = \cos t

\displaystyle \therefore \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\cos t}{-\sin t} = -\cot t

\displaystyle \text{Now, }\left(\frac{dy}{dx}\right)_{t=\frac{2\pi}{3}} = -\cot\left(\frac{2\pi}{3}\right) = \frac{1}{\sqrt{3}}

\displaystyle \textbf{Question 17: }~\text{If }x=a\!\left(t+\frac{1}{t}\right)\text{ and }y=a\!\left(t-\frac{1}{t}\right),\text{ prove that }\frac{dy}{dx}=\frac{x}{y}
\displaystyle \text{Answer:}

\displaystyle \text{We have, } x = a\left(t + \frac{1}{t}\right) \text{ and } y = a\left(t - \frac{1}{t}\right)

\displaystyle \Rightarrow \frac{dx}{dt} = a\frac{d}{dt}\left(t + \frac{1}{t}\right) \text{ and } \frac{dy}{dt} = a\frac{d}{dt}\left(t - \frac{1}{t}\right)

\displaystyle \Rightarrow \frac{dx}{dt} = a\left(1 - \frac{1}{t^{2}}\right) \text{ and } \frac{dy}{dt} = a\left(1 + \frac{1}{t^{2}}\right)

\displaystyle \Rightarrow \frac{dx}{dt} = a\left(\frac{t^{2}-1}{t^{2}}\right) \text{ and } \frac{dy}{dt} = a\left(\frac{t^{2}+1}{t^{2}}\right)

\displaystyle \therefore \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{a(t^{2}+1)}{t^{2}} \times \frac{t^{2}}{a(t^{2}-1)}

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

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{a\left(t+\frac{1}{t}\right)}{a\left(t-\frac{1}{t}\right)}

\displaystyle \therefore \frac{dy}{dx} = \frac{x}{y}

\displaystyle \textbf{Question 18: }~\text{If }x=\sin^{-1}\!\left(\frac{2t}{1+t^2}\right)\text{ and }y=\tan^{-1}\!\left(\frac{2t}{1-t^2}\right),\;-1<t<1,\text{ prove that }\frac{dy}{dx}=1
\displaystyle \text{Answer:}

\displaystyle \text{We have, } x = \sin^{-1}\left(\frac{2t}{1+t^{2}}\right)

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

\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 \therefore x = \sin^{-1}\left(\frac{2\tan\theta}{1+\tan^{2}\theta}\right) = \sin^{-1}(\sin 2\theta)

\displaystyle \Rightarrow x = 2\theta \qquad \left[\because -\frac{\pi}{2} < 2\theta < \frac{\pi}{2}\right]

\displaystyle \Rightarrow x = 2\tan^{-1} t \qquad \left[\because t = \tan\theta\right]

\displaystyle \Rightarrow \frac{dx}{dt} = \frac{2}{1+t^{2}} \ \ldots\ (i)

\displaystyle \text{Now, } y = \tan^{-1}\left(\frac{2t}{1-t^{2}}\right)

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

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

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

\displaystyle \Rightarrow y = 2\theta \qquad \left[\because -\frac{\pi}{2} < 2\theta < \frac{\pi}{2}\right]

\displaystyle \Rightarrow y = 2\tan^{-1} t \qquad \left[\because t = \tan\theta\right]

\displaystyle \Rightarrow \frac{dy}{dt} = \frac{2}{1+t^{2}} \ \ldots\ (ii)

\displaystyle \text{Dividing equation (ii) by (i),}

\displaystyle \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2}{1+t^{2}}\times\frac{1+t^{2}}{2}

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

\displaystyle \textbf{Question 19: }~\text{If }x=\frac{\sin^3 t}{\sqrt{\cos2t}},\;y=\frac{\cos^3 t}{\sqrt{\cos2t}},\text{ find }\frac{dy}{dx}
\displaystyle \text{Answer:}

\displaystyle \text{We have, } x=\frac{\sin^{3}t}{\sqrt{\cos 2t}} \text{ and } y=\frac{\cos^{3}t}{\sqrt{\cos 2t}}

\displaystyle \Rightarrow \frac{dx}{dt} = \frac{d}{dt}\!\left(\frac{\sin^{3}t}{\sqrt{\cos 2t}}\right)

\displaystyle \Rightarrow \frac{dx}{dt} = \frac{\sqrt{\cos 2t}\,\frac{d}{dt}(\sin^{3}t) -\sin^{3}t\,\frac{d}{dt}(\sqrt{\cos 2t})}{\cos 2t} \quad \text{[using quotient rule]}

\displaystyle \Rightarrow \frac{dx}{dt} = \frac{\sqrt{\cos 2t}(3\sin^{2}t)\cos t -\sin^{3}t\frac{1}{2\sqrt{\cos 2t}}\frac{d}{dt}(\cos 2t)}{\cos 2t}

\displaystyle \Rightarrow \frac{dx}{dt} = \frac{3\sqrt{\cos 2t}\sin^{2}t\cos t +\frac{\sin^{3}t}{2\sqrt{\cos 2t}}(2\sin 2t)}{\cos 2t}

\displaystyle \Rightarrow \frac{dx}{dt} = \frac{3\cos 2t\sin^{2}t\cos t+\sin^{3}t\sin 2t} {\cos 2t\sqrt{\cos 2t}}

\displaystyle \text{Now, } \frac{dy}{dt} = \frac{d}{dt}\!\left(\frac{\cos^{3}t}{\sqrt{\cos 2t}}\right)

\displaystyle \Rightarrow \frac{dy}{dt} = \frac{\sqrt{\cos 2t}\,\frac{d}{dt}(\cos^{3}t) -\cos^{3}t\,\frac{d}{dt}(\sqrt{\cos 2t})}{\cos 2t} \quad \text{[using quotient rule]}

\displaystyle \Rightarrow \frac{dy}{dt} = \frac{\sqrt{\cos 2t}(3\cos^{2}t)(-\sin t) -\cos^{3}t\frac{1}{2\sqrt{\cos 2t}}\frac{d}{dt}(\cos 2t)}{\cos 2t}

\displaystyle \Rightarrow \frac{dy}{dt} = \frac{-3\sqrt{\cos 2t}\cos^{2}t\sin t +\frac{\cos^{3}t}{2\sqrt{\cos 2t}}(2\sin 2t)}{\cos 2t}

\displaystyle \Rightarrow \frac{dy}{dt} = \frac{-3\cos 2t\cos^{2}t\sin t+\cos^{3}t\sin 2t} {\cos 2t\sqrt{\cos 2t}}

\displaystyle \therefore \frac{dy}{dx} = \frac{-3\cos 2t\cos^{2}t\sin t+\cos^{3}t\sin 2t} {3\cos 2t\sin^{2}t\cos t+\sin^{3}t\sin 2t}

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{\sin t\cos t[-3\cos 2t\cos t+2\cos^{3}t]} {\sin t\cos t[3\cos 2t\sin t+2\sin^{3}t]}

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{-3(2\cos^{2}t-1)\cos t+2\cos^{3}t} {3(1-2\sin^{2}t)\sin t+2\sin^{3}t}

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{-4\cos^{3}t+3\cos t}{3\sin t-4\sin^{3}t}

\displaystyle \Rightarrow \frac{dy}{dx} = -\frac{\cos 3t}{\sin 3t}

\displaystyle \therefore \frac{dy}{dx} = -\cot 3t

\displaystyle \textbf{Question 20: }~\text{If }x=\left(t+\frac{1}{t}\right)^a,\;y=a^{\,t+\frac{1}{t}},\text{ find }\frac{dy}{dx}
\displaystyle \text{Answer:}

\displaystyle \text{We have, } x=\left(t+\frac{1}{t}\right)^{a}

\displaystyle \Rightarrow \frac{dx}{dt} = \frac{d}{dt}\left[\left(t+\frac{1}{t}\right)^{a}\right]

\displaystyle \Rightarrow \frac{dx}{dt} = a\left(t+\frac{1}{t}\right)^{a-1} \frac{d}{dt}\left(t+\frac{1}{t}\right)

\displaystyle \Rightarrow \frac{dx}{dt} = a\left(t+\frac{1}{t}\right)^{a-1} \left(1-\frac{1}{t^{2}}\right) \quad \ldots (i)

\displaystyle \text{and,}

\displaystyle y = a^{\left(t+\frac{1}{t}\right)}

\displaystyle \Rightarrow \frac{dy}{dt} = \frac{d}{dt}\left[a^{\left(t+\frac{1}{t}\right)}\right]

\displaystyle \Rightarrow \frac{dy}{dt} = a^{\left(t+\frac{1}{t}\right)}\log a \frac{d}{dt}\left(t+\frac{1}{t}\right)

\displaystyle \Rightarrow \frac{dy}{dt} = a^{\left(t+\frac{1}{t}\right)}\log a \left(1-\frac{1}{t^{2}}\right) \quad \ldots (ii)

\displaystyle \text{Dividing equation (ii) by (i),}

\displaystyle \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{a^{\left(t+\frac{1}{t}\right)}\log a \left(1-\frac{1}{t^{2}}\right)} {a\left(t+\frac{1}{t}\right)^{a-1} \left(1-\frac{1}{t^{2}}\right)}

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{a^{\left(t+\frac{1}{t}\right)}\log a} {a\left(t+\frac{1}{t}\right)^{a-1}}

\displaystyle \textbf{Question 21: }~\text{If }x=a\!\left(\frac{1+t^2}{1-t^2}\right)\text{ and }y=\frac{2t}{1-t^2},\text{ find }\frac{dy}{dx}
\displaystyle \text{Answer:}

\displaystyle \text{We have, } x = a\left(\frac{1+t^{2}}{1-t^{2}}\right)

\displaystyle \Rightarrow \frac{dx}{dt} = a\left[\frac{(1-t^{2})\frac{d}{dt}(1+t^{2})-(1+t^{2})\frac{d}{dt}(1-t^{2})}{(1-t^{2})^{2}}\right] \ \text{[using quotient rule]}

\displaystyle \Rightarrow \frac{dx}{dt} = a\left[\frac{(1-t^{2})(2t)-(1+t^{2})(-2t)}{(1-t^{2})^{2}}\right]

\displaystyle \Rightarrow \frac{dx}{dt} = a\left[\frac{2t-2t^{3}+2t+2t^{3}}{(1-t^{2})^{2}}\right]

\displaystyle \Rightarrow \frac{dx}{dt} = \frac{4at}{(1-t^{2})^{2}} \ \ldots (i)

\displaystyle \text{and,}

\displaystyle y = \frac{2t}{1-t^{2}}

\displaystyle \Rightarrow \frac{dy}{dt} = 2\left[\frac{(1-t^{2})\frac{d}{dt}(t)-t\frac{d}{dt}(1-t^{2})}{(1-t^{2})^{2}}\right] \ \text{[using quotient rule]}

\displaystyle \Rightarrow \frac{dy}{dt} = 2\left[\frac{(1-t^{2})(1)-t(-2t)}{(1-t^{2})^{2}}\right]

\displaystyle \Rightarrow \frac{dy}{dt} = 2\left[\frac{1-t^{2}+2t^{2}}{(1-t^{2})^{2}}\right]

\displaystyle \Rightarrow \frac{dy}{dt} = \frac{2(1+t^{2})}{(1-t^{2})^{2}} \ \ldots (ii)

\displaystyle \text{Dividing equation (ii) by (i),}

\displaystyle \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2(1+t^{2})}{(1-t^{2})^{2}} \times \frac{(1-t^{2})^{2}}{4at}

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{1+t^{2}}{2at}

\displaystyle \textbf{Question 22: }~\text{If }x=10(t-\sin t)\text{ and }y=12(1-\cos t),\text{ find }\frac{dy}{dx}\;
\displaystyle \text{Answer:}

\displaystyle \text{We have, } x = 10(t-\sin t) \text{ and } y = 12(1-\cos t)

\displaystyle \Rightarrow \frac{dx}{dt} = \frac{d}{dt}[10(t-\sin t)] \text{ and } \frac{dy}{dt} = \frac{d}{dt}[12(1-\cos t)]

\displaystyle \Rightarrow \frac{dx}{dt} = 10\frac{d}{dt}(t-\sin t) \text{ and } \frac{dy}{dt} = 12\frac{d}{dt}(1-\cos t)

\displaystyle \Rightarrow \frac{dx}{dt} = 10(1-\cos t) \text{ and } \frac{dy}{dt} = 12\sin t

\displaystyle \therefore \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{12\sin t}{10(1-\cos t)}

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{12\cdot 2\sin\frac{t}{2}\cos\frac{t}{2}} {10\cdot 2\sin^{2}\frac{t}{2}}

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{6}{5}\cot\frac{t}{2}

\displaystyle \textbf{Question 23: }~\text{If }x=a(\theta-\sin\theta)\text{ and }y=a(1+\cos\theta),\text{ find }\frac{dy}{dx}\text{ at }\theta=\frac{\pi}{3}\; \hspace{5.0cm} [\text{CBSE 2011}]
\displaystyle \text{Answer:}

\displaystyle \text{We have, } x = a(\theta-\sin\theta) \text{ and } y = a(1+\cos\theta)

\displaystyle \Rightarrow \frac{dx}{d\theta} = \frac{d}{d\theta}[a(\theta-\sin\theta)] \text{ and } \frac{dy}{d\theta} = \frac{d}{d\theta}[a(1+\cos\theta)]

\displaystyle \Rightarrow \frac{dx}{d\theta} = a(1-\cos\theta) \text{ and } \frac{dy}{d\theta} = -a\sin\theta

\displaystyle \therefore \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{-a\sin\theta}{a(1-\cos\theta)}

\displaystyle \text{Now, } \left[\frac{dy}{dx}\right]_{\theta=\frac{\pi}{3}} = -\frac{\sin\frac{\pi}{3}}{1-\cos\frac{\pi}{3}} = -\frac{\frac{\sqrt{3}}{2}}{1-\frac{1}{2}} = -\sqrt{3}

\displaystyle \textbf{Question 24: }~\text{If }x=a\sin2t(1+\cos2t)\text{ and }y=b\cos2t(1-\cos2t),\text{ show that at }t=\frac{\pi}{4},\;\frac{dy}{dx}=\frac{b}{a}\; \hspace{5.0cm} [\text{CBSE 2014, 2016}]
\displaystyle \text{Answer:}

\displaystyle x = a\sin 2t(1+\cos 2t) \text{ and } y = b\cos 2t(1-\cos 2t)

\displaystyle \Rightarrow \frac{dx}{dt} = 2a\cos 2t(1+\cos 2t) + 2a\sin 2t(1-\cos 2t) \text{ and } \frac{dy}{dt} = -2b\sin 2t(1-\cos 2t) + 2b\cos 2t(1+\cos 2t)

\displaystyle \Rightarrow \frac{dx}{dt} = 2a(\cos 2t+\cos^{2}2t+\sin 2t-\sin 2t\cos 2t) \text{ and } \frac{dy}{dt} = 2b(-\sin 2t+\sin 2t\cos 2t+\cos 2t+\cos^{2}2t)

\displaystyle \therefore \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-2b(-\sin 2t+\sin 2t\cos 2t+\cos 2t+\cos^{2}2t)} {2a(\cos 2t+\cos^{2}2t+\sin 2t-\sin 2t\cos 2t)}

\displaystyle \Rightarrow \left(\frac{dy}{dx}\right)_{t=\frac{\pi}{4}} = \frac{-2b\left(-\sin\frac{\pi}{2} +\sin\frac{\pi}{2}\cos\frac{\pi}{2} +\cos\frac{\pi}{2} +\cos^{2}\frac{\pi}{2}\right)} {2a\left(\cos\frac{\pi}{2} +\cos^{2}\frac{\pi}{2} +\sin\frac{\pi}{2} -\sin\frac{\pi}{2}\cos\frac{\pi}{2}\right)} = \frac{b}{a}

\displaystyle \textbf{Question 25: }~\text{If }x=\cos t(3-2\cos^2 t)\text{ and }y=\sin t(3-2\sin^2 t),\text{ find the value of }\frac{dy}{dx}\text{ at }t=\frac{\pi}{4}\; \hspace{5.0cm} [\text{CBSE 2014}]
\displaystyle \text{Answer:}

\displaystyle \text{We have, } x=\cos t\,(3-2\cos^{2}t) \text{ and } y=\sin t\,(3-2\sin^{2}t)

\displaystyle \Rightarrow \frac{dx}{dt} = -\sin t(3-2\cos^{2}t) +\cos t\,[4\cos t\sin t]

\displaystyle \Rightarrow \frac{dy}{dt} = \cos t(3-2\sin^{2}t) +\sin t[-4\sin t\cos t]

\displaystyle \Rightarrow \frac{dx}{dt} = -3\sin t+6\sin t\cos^{2}t

\displaystyle \Rightarrow \frac{dy}{dt} = 3\cos t-6\sin^{2}t\cos t

\displaystyle \Rightarrow \frac{dx}{dt} = -3\sin t(1-2\cos^{2}t) \text{ and } \frac{dy}{dt} = 3\cos t(1-2\sin^{2}t)

\displaystyle \Rightarrow \frac{dx}{dt} = 3\sin t\cos 2t \text{ and } \frac{dy}{dt} = 3\cos t\cos 2t

\displaystyle \therefore \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{3\cos t\cos 2t}{3\sin t\cos 2t} = \cot t

\displaystyle \text{Now, } \left(\frac{dy}{dx}\right)_{t=\frac{\pi}{4}} = \cot\frac{\pi}{4} = 1

\displaystyle \textbf{Question 26: }~\text{If }x=\frac{1+\log t}{t^2}\text{ and }y=\frac{3+2\log t}{t},\text{ find }\frac{dy}{dx}\;
\displaystyle \text{Answer:}

\displaystyle \text{We have, } x=\frac{1+\log t}{t^{2}} \text{ and } y=\frac{3+2\log t}{t}

\displaystyle \Rightarrow \frac{dx}{dt} =\frac{t^2\frac{d}{dt}(1+\log t)-(1+\log t)\frac{d}{dt}(t^2)}{t^{4}}

\displaystyle \Rightarrow \frac{dx}{dt} =\frac{t^2\left(\frac{1}{t}\right)-2t(1+\log t)}{t^{4}}

\displaystyle \Rightarrow \frac{dx}{dt} =\frac{-1-2\log t}{t^{3}}

\displaystyle \Rightarrow \frac{dy}{dt} =\frac{t\frac{d}{dt}(3+2\log t)-(3+2\log t)\frac{d}{dt}(t)}{t^{2}}

\displaystyle \Rightarrow \frac{dy}{dt} =\frac{2-(3+2\log t)}{t^{2}}

\displaystyle \Rightarrow \frac{dy}{dt} =\frac{-1-2\log t}{t^{2}}

\displaystyle \therefore \frac{dy}{dx} =\frac{\frac{dy}{dt}}{\frac{dx}{dt}} =\frac{\frac{-1-2\log t}{t^{2}}}{\frac{-1-2\log t}{t^{3}}} =t

\displaystyle \textbf{Question 27: }~\text{If }x=3\sin t-\sin3t\text{ and }y=3\cos t-\cos3t,\text{ find }\frac{dy}{dx}\text{ at }t=\frac{\pi}{3}\;
\displaystyle \text{Answer:}

\displaystyle \sin x=\frac{2t}{1+t^{2}} \text{ and } \tan y=\frac{2t}{1-t^{2}}

\displaystyle \Rightarrow x=\sin^{-1}\!\left(\frac{2t}{1+t^{2}}\right) \text{ and } y=\tan^{-1}\!\left(\frac{2t}{1-t^{2}}\right)

\displaystyle \Rightarrow x=2\tan^{-1}t \text{ and } y=2\tan^{-1}t

\displaystyle \Rightarrow \frac{dx}{dt} =\frac{2}{1+t^{2}} \text{ and } \frac{dy}{dt} =\frac{2}{1+t^{2}}

\displaystyle \therefore \frac{dy}{dx} =\frac{\frac{dy}{dt}}{\frac{dx}{dt}} =\frac{\frac{2}{1+t^{2}}}{\frac{2}{1+t^{2}}} =1

\displaystyle \textbf{Question 28: }~\text{If }\sin x=\frac{2t}{1+t^2}\text{ and }\tan y=\frac{2t}{1-t^2},\text{ find }\frac{dy}{dx}\;
\displaystyle \text{Answer:}

\displaystyle \text{Let } u=\sin x \text{ and } v=\cos x

\displaystyle \Rightarrow \frac{du}{dx}=\cos x \text{ and } \frac{dv}{dx}=-\sin x

\displaystyle \therefore \frac{du}{dv} =\frac{\frac{du}{dx}}{\frac{dv}{dx}} =\frac{\cos x}{-\sin x}

\displaystyle \Rightarrow \frac{du}{dv}=-\cot x

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