\displaystyle \text{Differentiate the following functions from first principles:}

\displaystyle \text{Question 1: } e^{-x}

\displaystyle \text{Answer:}

\displaystyle \text{Let } f(x)=e^{-x}.

\displaystyle \Rightarrow f(x+h)=e^{-(x+h)}

\displaystyle \frac{d}{dx}\{f(x)\} =\lim_{h\to0}\frac{f(x+h)-f(x)}{h}

\displaystyle =\lim_{h\to0}\frac{e^{-(x+h)}-e^{-x}}{h}

\displaystyle =\lim_{h\to0}\frac{e^{-x}e^{-h}-e^{-x}}{h}

\displaystyle =\lim_{h\to0}e^{-x}\frac{e^{-h}-1}{h}

\displaystyle =-e^{-x}\lim_{h\to0}\frac{e^{-h}-1}{-h}

\displaystyle =-e^{-x}\left[\because\ \lim_{h\to0}\frac{e^{-h}-1}{-h}=1\right]

\displaystyle \therefore\ \frac{d}{dx}\left(e^{-x}\right)=-e^{-x}.

\displaystyle \text{Question 2: } e^{3x}

\displaystyle \text{Answer:}

\displaystyle \text{Let } f(x)=e^{3x}.

\displaystyle \Rightarrow f(x+h)=e^{3(x+h)}

\displaystyle \frac{d}{dx}\{f(x)\} =\lim_{h\to0}\frac{f(x+h)-f(x)}{h}

\displaystyle =\lim_{h\to0}\frac{e^{3(x+h)}-e^{3x}}{h}

\displaystyle =\lim_{h\to0}\frac{e^{3x}e^{3h}-e^{3x}}{h}

\displaystyle =\lim_{h\to0}e^{3x}\frac{e^{3h}-1}{h}

\displaystyle =\lim_{h\to0}e^{3x}\cdot 3\frac{e^{3h}-1}{3h}

\displaystyle =3e^{3x}\lim_{h\to0}\frac{e^{3h}-1}{3h}

\displaystyle =3e^{3x}\left[\because\ \lim_{h\to0}\frac{e^{3h}-1}{3h}=1\right]

\displaystyle \therefore\ \frac{d}{dx}\left(e^{3x}\right)=3e^{3x}.

\displaystyle \text{Question 3: } e^{ax+b}

\displaystyle \text{Answer:}

\displaystyle \text{Let } f(x)=e^{ax+b}.

\displaystyle \Rightarrow f(x+h)=e^{a(x+h)+b}

\displaystyle \therefore\ \frac{d}{dx}\{f(x)\} =\lim_{h\to0}\frac{f(x+h)-f(x)}{h}

\displaystyle =\lim_{h\to0}\frac{e^{a(x+h)+b}-e^{ax+b}}{h}

\displaystyle =\lim_{h\to0}\frac{e^{ax+b}e^{ah}-e^{ax+b}}{h}

\displaystyle =\lim_{h\to0}e^{ax+b}\frac{e^{ah}-1}{h}

\displaystyle =\lim_{h\to0}e^{ax+b}\cdot a\frac{e^{ah}-1}{ah}

\displaystyle =a e^{ax+b}\lim_{h\to0}\frac{e^{ah}-1}{ah}

\displaystyle =a e^{ax+b}\left[\because\ \lim_{h\to0}\frac{e^{ah}-1}{ah}=1\right]

\displaystyle \therefore\ \frac{d}{dx}\left(e^{ax+b}\right)=a e^{ax+b}.

\displaystyle \text{Question 4: } e^{\cos x}

\displaystyle \text{Answer:}

\displaystyle \text{Let } f(x)=e^{\cos x}.

\displaystyle \Rightarrow f(x+h)=e^{\cos(x+h)}

\displaystyle \therefore\ \frac{d}{dx}\{f(x)\} =\lim_{h\to0}\frac{f(x+h)-f(x)}{h}

\displaystyle =\lim_{h\to0}\frac{e^{\cos(x+h)}-e^{\cos x}}{h}

\displaystyle =\lim_{h\to0}e^{\cos x}\left[\frac{e^{\cos(x+h)-\cos x}-1}{h}\right]

\displaystyle =\lim_{h\to0}e^{\cos x} \left(\frac{\cos(x+h)-\cos x}{h}\right) \left[\frac{e^{\cos(x+h)-\cos x}-1}{\cos(x+h)-\cos x}\right]

\displaystyle =e^{\cos x} \left[\lim_{h\to0}\frac{\cos(x+h)-\cos x}{h}\right] \left[\because\ \lim_{u\to0}\frac{e^{u}-1}{u}=1\right]

\displaystyle =e^{\cos x} \left[\lim_{h\to0}\frac{-2\sin\!\left(\frac{2x+h}{2}\right)\sin\!\left(\frac{h}{2}\right)}{h}\right]

\displaystyle =e^{\cos x} \left[-\sin\!\left(\frac{2x+h}{2}\right)\cdot \lim_{h\to0}\frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}}\right]

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

\displaystyle \therefore\ \frac{d}{dx}\left(e^{\cos x}\right)=-\sin x\,e^{\cos x}.

\displaystyle \text{Question 5: } e^{\sqrt{2x}}

\displaystyle \text{Answer:}

\displaystyle \text{Let } f(x)=e^{\sqrt{2x}}.

\displaystyle \Rightarrow f(x+h)=e^{\sqrt{2(x+h)}}

\displaystyle \therefore\ \frac{d}{dx}\{f(x)\} =\lim_{h\to0}\frac{f(x+h)-f(x)}{h}

\displaystyle =\lim_{h\to0}\frac{e^{\sqrt{2(x+h)}}-e^{\sqrt{2x}}}{h}

\displaystyle =\lim_{h\to0}e^{\sqrt{2x}} \left[\frac{e^{\sqrt{2(x+h)}-\sqrt{2x}}-1}{h}\right]

\displaystyle =\lim_{h\to0}e^{\sqrt{2x}} \left(\frac{\sqrt{2(x+h)}-\sqrt{2x}}{h}\right) \left[\frac{e^{\sqrt{2(x+h)}-\sqrt{2x}}-1}{\sqrt{2(x+h)}-\sqrt{2x}}\right]

\displaystyle =e^{\sqrt{2x}} \left[\lim_{h\to0}\frac{\sqrt{2(x+h)}-\sqrt{2x}}{h}\right] \left[\because\ \lim_{u\to0}\frac{e^{u}-1}{u}=1\right]

\displaystyle =e^{\sqrt{2x}} \lim_{h\to0}\frac{\sqrt{2(x+h)}-\sqrt{2x}}{h} \cdot \frac{\sqrt{2(x+h)}+\sqrt{2x}}{\sqrt{2(x+h)}+\sqrt{2x}}

\displaystyle =e^{\sqrt{2x}} \lim_{h\to0}\frac{2(x+h)-2x}{h\left(\sqrt{2(x+h)}+\sqrt{2x}\right)}

\displaystyle =e^{\sqrt{2x}} \lim_{h\to0}\frac{2h}{h\left(\sqrt{2(x+h)}+\sqrt{2x}\right)}

\displaystyle =e^{\sqrt{2x}}\cdot\frac{2}{2\sqrt{2x}}

\displaystyle =\frac{e^{\sqrt{2x}}}{\sqrt{2x}}

\displaystyle \therefore\ \frac{d}{dx}\left(e^{\sqrt{2x}}\right)=\frac{e^{\sqrt{2x}}}{\sqrt{2x}}.

\displaystyle \text{Differentiate each of the following functions from first principles:}

\displaystyle \text{Question 6: } \log \cos x

\displaystyle \text{Answer:}

\displaystyle \text{Let } f(x)=\log(\cos x).

\displaystyle \Rightarrow f(x+h)=\log(\cos(x+h))

\displaystyle \therefore\ \frac{d}{dx}\{f(x)\} =\lim_{h\to0}\frac{f(x+h)-f(x)}{h}

\displaystyle =\lim_{h\to0}\frac{\log(\cos(x+h))-\log(\cos x)}{h}

\displaystyle =\lim_{h\to0}\frac{\log\!\left(\dfrac{\cos(x+h)}{\cos x}\right)}{h} \quad [\because\ \log A-\log B=\log(A/B)]

\displaystyle =\lim_{h\to0}\frac{\log\!\left(1+\dfrac{\cos(x+h)-\cos x}{\cos x}\right)}{h}

\displaystyle =\lim_{h\to0} \left[\frac{\log\!\left(1+\dfrac{\cos(x+h)-\cos x}{\cos x}\right)} {\dfrac{\cos(x+h)-\cos x}{\cos x}}\right] \left[\frac{\cos(x+h)-\cos x}{h}\right]

\displaystyle =\left[\because\ \lim_{u\to0}\frac{\log(1+u)}{u}=1\right] \lim_{h\to0}\frac{\cos(x+h)-\cos x}{h\cos x}

\displaystyle =\lim_{h\to0} \frac{-2\sin\!\left(\frac{(x+h)+x}{2}\right)\sin\!\left(\frac{(x+h)-x}{2}\right)} {h\cos x}

\displaystyle =\lim_{h\to0} \frac{-2\sin\!\left(\frac{2x+h}{2}\right)\sin\!\left(\frac{h}{2}\right)} {h\cos x}

\displaystyle =-\frac{\sin x}{\cos x} \left[\because\ \lim_{h\to0}\frac{\sin(h/2)}{h/2}=1\right]

\displaystyle =-\tan x

\displaystyle \therefore\ \frac{d}{dx}\big(\log(\cos x)\big)=-\tan x.

\displaystyle \text{Question 7: } e^{\sqrt{ \cot x}}

\displaystyle \text{Answer:}

\displaystyle \text{Let } f(x)=e^{\sqrt{\cot x}}.

\displaystyle \Rightarrow f(x+h)=e^{\sqrt{\cot(x+h)}}

\displaystyle \therefore\ \frac{d}{dx}\{f(x)\} =\lim_{h\to0}\frac{f(x+h)-f(x)}{h}

\displaystyle =\lim_{h\to0}\frac{e^{\sqrt{\cot(x+h)}}-e^{\sqrt{\cot x}}}{h}

\displaystyle =\lim_{h\to0}e^{\sqrt{\cot x}} \left[\frac{e^{\sqrt{\cot(x+h)}-\sqrt{\cot x}}-1}{h}\right]

\displaystyle =\lim_{h\to0}e^{\sqrt{\cot x}} \left(\frac{\sqrt{\cot(x+h)}-\sqrt{\cot x}}{h}\right) \left[\frac{e^{\sqrt{\cot(x+h)}-\sqrt{\cot x}}-1} {\sqrt{\cot(x+h)}-\sqrt{\cot x}}\right]

\displaystyle =e^{\sqrt{\cot x}} \left[\lim_{h\to0}\frac{\sqrt{\cot(x+h)}-\sqrt{\cot x}}{h}\right] \left[\because\ \lim_{u\to0}\frac{e^{u}-1}{u}=1\right]

\displaystyle =e^{\sqrt{\cot x}} \lim_{h\to0} \frac{\cot(x+h)-\cot x} {h\big(\sqrt{\cot(x+h)}+\sqrt{\cot x}\big)}

\displaystyle =e^{\sqrt{\cot x}} \left[\lim_{h\to0}\frac{-\mathrm{cosec}^2 x}{\sqrt{\cot x}+\sqrt{\cot x}}\right]

\displaystyle =-\,\frac{e^{\sqrt{\cot x}}\mathrm{cosec}^2 x}{2\sqrt{\cot x}}

\displaystyle \therefore\ \frac{d}{dx}\!\left(e^{\sqrt{\cot x}}\right) =-\frac{e^{\sqrt{\cot x}}\mathrm{cosec}^2 x}{2\sqrt{\cot x}}.

\displaystyle \text{Question 8: } x^2 e^x

\displaystyle \text{Answer:}

\displaystyle \text{Let } f(x)=x^{2}e^{x}.

\displaystyle \Rightarrow f(x+h)=(x+h)^{2}e^{x+h}

\displaystyle \therefore\ \frac{d}{dx}\{f(x)\} =\lim_{h\to0}\frac{f(x+h)-f(x)}{h}

\displaystyle =\lim_{h\to0}\frac{(x+h)^{2}e^{x+h}-x^{2}e^{x}}{h}

\displaystyle =\lim_{h\to0}\frac{(x^{2}+2xh+h^{2})e^{x+h}-x^{2}e^{x}}{h}

\displaystyle =\lim_{h\to0}\left[ \frac{x^{2}e^{x+h}-x^{2}e^{x}}{h} +\frac{2xh\,e^{x+h}}{h} +\frac{h^{2}e^{x+h}}{h} \right]

\displaystyle =\lim_{h\to0}\left[ x^{2}e^{x}\frac{e^{h}-1}{h} +2x\,e^{x+h} +h\,e^{x+h} \right]

\displaystyle =x^{2}e^{x}(1)+2x e^{x}+0 \quad\left[\because\ \lim_{h\to0}\frac{e^{h}-1}{h}=1\right]

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

\displaystyle \therefore\ \frac{d}{dx}\left(x^{2}e^{x}\right)=e^{x}(x^{2}+2x).

\displaystyle \text{Question 9: } \log \mathrm{cosec} x

\displaystyle \text{Answer:}

\displaystyle \text{Let } f(x)=\log(\mathrm{cosec} x).

\displaystyle \Rightarrow f(x+h)=\log(\mathrm{cosec} (x+h))

\displaystyle \therefore\ \frac{d}{dx}\{f(x)\} =\lim_{h\to0}\frac{f(x+h)-f(x)}{h}

\displaystyle =\lim_{h\to0}\frac{\log(\mathrm{cosec}(x+h))-\log(\mathrm{cosec} x)}{h}

\displaystyle =\lim_{h\to0}\frac{\log\left(\dfrac{\mathrm{cosec}(x+h)}{\mathrm{cosec} x}\right)}{h}

\displaystyle =\lim_{h\to0}\frac{\log\left(\dfrac{\sin x}{\sin(x+h)}\right)}{h}

\displaystyle =\lim_{h\to0}\frac{\log\left(1+\dfrac{\sin x-\sin(x+h)}{\sin(x+h)}\right)}{h}

\displaystyle =\lim_{h\to0} \left[ \frac{\log\left(1+\dfrac{\sin x-\sin(x+h)}{\sin(x+h)}\right)} {\dfrac{\sin x-\sin(x+h)}{\sin(x+h)}} \right] \cdot \lim_{h\to0} \frac{\sin x-\sin(x+h)}{h}

\displaystyle =\lim_{h\to0}\frac{\sin x-\sin(x+h)}{h} \quad \left[\because\ \lim_{u\to0}\frac{\log(1+u)}{u}=1\right]

\displaystyle =\lim_{h\to0} \frac{-2\cos\left(\frac{2x+h}{2}\right)\sin\left(\frac{h}{2}\right)}{h} \quad \left[\because\ \sin A-\sin B=-2\cos\frac{A+B}{2}\sin\frac{A-B}{2}\right]

\displaystyle =-\cos x\cdot\lim_{h\to0}\frac{2\sin\left(\frac{h}{2}\right)}{h}

\displaystyle =-\cos x\cdot 1 \quad \left[\because\ \lim_{h\to0}\frac{\sin(h/2)}{h/2}=1\right]

\displaystyle =-\cot x

\displaystyle \therefore\ \frac{d}{dx}(\log\mathrm{cosec} x)=-\cot x.

\displaystyle \text{Question 10: } \sin^{-1} (2x+3)

\displaystyle \text{Answer:}

\displaystyle \text{Given:}

\displaystyle f(x)=\sin^{-1}(2x+3)

\displaystyle \text{We find the derivative of } f(x) \text{ using first principles.}

\displaystyle f(x+h)=\sin^{-1}(2x+2h+3)

\displaystyle \frac{d}{dx}\{f(x)\} = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}

\displaystyle = \lim_{h\to 0}\frac{\sin^{-1}(2x+2h+3)-\sin^{-1}(2x+3)}{h}

\displaystyle = \lim_{h\to 0}\frac{1}{h} \left[ \sin^{-1}\!\left( (2x+2h+3)\sqrt{1-(2x+3)^2} -(2x+3)\sqrt{1-(2x+2h+3)^2} \right) \right]

\displaystyle = \lim_{h\to 0}\frac{z}{h}, \quad \text{where } z=(2x+2h+3)\sqrt{1-(2x+3)^2} -(2x+3)\sqrt{1-(2x+2h+3)^2}

\displaystyle = \lim_{h\to 0} \frac{(2x+2h+3)^2\{1-(2x+3)^2\} -(2x+3)^2\{1-(2x+2h+3)^2\}} {h\left[(2x+2h+3)\sqrt{1-(2x+3)^2} +(2x+3)\sqrt{1-(2x+2h+3)^2}\right]}

\displaystyle = \lim_{h\to 0} \frac{4h(2x+3)} {2(2x+3)\sqrt{1-(2x+3)^2}}

\displaystyle = \frac{2}{\sqrt{1-(2x+3)^2}}

\displaystyle \therefore\ \frac{d}{dx}\big(\sin^{-1}(2x+3)\big) = \frac{2}{\sqrt{1-(2x+3)^2}}

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