Class 11: Derivatives – Exercise 30.2

$\displaystyle \textbf{Question 1: } \text{Differentiate each of the following from first principles:}$
$\displaystyle \text{(i) }\frac{2}{x}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=\frac{2}{x}.\text{ Then }f(x+h)=\frac{2}{x+h}.$
$\displaystyle \therefore \frac{d}{dx}(f(x))=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{\frac{2}{x+h}-\frac{2}{x}}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{2x-2(x+h)}{hx(x+h)}$
$\displaystyle =\lim\limits_{h\to0}\frac{-2h}{hx(x+h)}$
$\displaystyle =\lim\limits_{h\to0}\frac{-2}{x(x+h)}$
$\displaystyle =-\frac{2}{x^2}$
$\displaystyle \text{Hence, }\frac{d}{dx}\left(\frac{2}{x}\right)=-\frac{2}{x^2}.$

$\displaystyle \text{(ii) }\frac{1}{\sqrt{x}}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=\frac{1}{\sqrt{x}}.\text{ Then }f(x+h)=\frac{1}{\sqrt{x+h}}.$
$\displaystyle \therefore \frac{d}{dx}(f(x))=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{\sqrt{x}-\sqrt{x+h}}{h\sqrt{x}\sqrt{x+h}}\times\frac{\sqrt{x}+\sqrt{x+h}}{\sqrt{x}+\sqrt{x+h}}$
$\displaystyle =\lim\limits_{h\to0}\frac{x-(x+h)}{(h\sqrt{x}\sqrt{x+h})(\sqrt{x}+\sqrt{x+h})}$
$\displaystyle =\lim\limits_{h\to0}\frac{-h}{(h\sqrt{x}\sqrt{x+h})(\sqrt{x}+\sqrt{x+h})}$
$\displaystyle =\lim\limits_{h\to0}\frac{-1}{(\sqrt{x}\sqrt{x+h})(\sqrt{x}+\sqrt{x+h})}$
$\displaystyle =-\frac{1}{\sqrt{x}\sqrt{x}(\sqrt{x}+\sqrt{x})}$
$\displaystyle =-\frac{1}{2x\sqrt{x}}$
$\displaystyle \text{Hence, }\frac{d}{dx}\left(\frac{1}{\sqrt{x}}\right)=-\frac{1}{2x\sqrt{x}}.$

$\displaystyle \text{(iii) }\frac{1}{x^3}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=\frac{1}{x^3}.\text{ Then }f(x+h)=\frac{1}{(x+h)^3}.$
$\displaystyle \therefore \frac{d}{dx}(f(x))=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{\frac{1}{(x+h)^3}-\frac{1}{x^3}}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{x^3-(x+h)^3}{h(x+h)^3x^3}$
$\displaystyle =\lim\limits_{h\to0}\frac{x^3-x^3-3x^2h-3xh^2-h^3}{h(x+h)^3x^3}$
$\displaystyle =\lim\limits_{h\to0}\frac{-3x^2h-3xh^2-h^3}{h(x+h)^3x^3}$
$\displaystyle =\lim\limits_{h\to0}\frac{h(-3x^2-3xh-h^2)}{h(x+h)^3x^3}$
$\displaystyle =\lim\limits_{h\to0}\frac{-3x^2-3xh-h^2}{(x+h)^3x^3}$
$\displaystyle =-\frac{3x^2}{x^6}=-\frac{3}{x^4}$
$\displaystyle \text{Hence, }\frac{d}{dx}\left(\frac{1}{x^3}\right)=-\frac{3}{x^4}.$

$\displaystyle \text{(iv) }\frac{x^2+1}{x}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=\frac{x^2+1}{x}.\text{ Then }f(x+h)=\frac{(x+h)^2+1}{x+h}.$
$\displaystyle \therefore \frac{d}{dx}(f(x))=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{\frac{(x+h)^2+1}{x+h}-\frac{x^2+1}{x}}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{x(x^2+2xh+h^2+1)-(x+h)(x^2+1)}{xh(x+h)}$
$\displaystyle =\lim\limits_{h\to0}\frac{x^3+2x^2h+xh^2+x-x^3-x^2h-x-h}{xh(x+h)}$
$\displaystyle =\lim\limits_{h\to0}\frac{x^2h+xh^2-h}{xh(x+h)}$
$\displaystyle =\lim\limits_{h\to0}\frac{h(x^2+xh-1)}{xh(x+h)}$
$\displaystyle =\lim\limits_{h\to0}\frac{x^2+xh-1}{x(x+h)}$
$\displaystyle =\frac{x^2-1}{x^2}$
$\displaystyle \text{Hence, }\frac{d}{dx}\left(\frac{x^2+1}{x}\right)=\frac{x^2-1}{x^2}.$

$\displaystyle \text{(v) }\frac{x^2-1}{x}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=\frac{x^2-1}{x}.\text{ Then }f(x+h)=\frac{(x+h)^2-1}{x+h}.$
$\displaystyle \therefore \frac{d}{dx}(f(x))=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{\frac{(x+h)^2-1}{x+h}-\frac{x^2-1}{x}}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{x(x^2+2xh+h^2-1)-(x+h)(x^2-1)}{xh(x+h)}$
$\displaystyle =\lim\limits_{h\to0}\frac{x^3+2x^2h+xh^2-x-x^3-x^2h+x+h}{xh(x+h)}$
$\displaystyle =\lim\limits_{h\to0}\frac{x^2h+xh^2+h}{xh(x+h)}$
$\displaystyle =\lim\limits_{h\to0}\frac{h(x^2+xh+1)}{xh(x+h)}$
$\displaystyle =\lim\limits_{h\to0}\frac{x^2+xh+1}{x(x+h)}$
$\displaystyle =\frac{x^2+1}{x^2}$
$\displaystyle \text{Hence, }\frac{d}{dx}\left(\frac{x^2-1}{x}\right)=\frac{x^2+1}{x^2}.$

$\displaystyle \text{(vi) }\frac{x+1}{x+2}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=\frac{x+1}{x+2}.\text{ Then }f(x+h)=\frac{x+h+1}{x+h+2}.$
$\displaystyle \therefore \frac{d}{dx}(f(x))=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{\frac{x+h+1}{x+h+2}-\frac{x+1}{x+2}}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{(x+h+1)(x+2)-(x+h+2)(x+1)}{h(x+h+2)(x+2)}$
$\displaystyle =\lim\limits_{h\to0}\frac{x^2+3x+2+hx+2h-x^2-3x-2-hx-h}{h(x+h+2)(x+2)}$
$\displaystyle =\lim\limits_{h\to0}\frac{h}{h(x+h+2)(x+2)}$
$\displaystyle =\lim\limits_{h\to0}\frac{1}{(x+h+2)(x+2)}$
$\displaystyle =\frac{1}{(x+2)^2}$
$\displaystyle \text{Hence, }\frac{d}{dx}\left(\frac{x+1}{x+2}\right)=\frac{1}{(x+2)^2}.$

$\displaystyle \text{(vii) }\frac{x+2}{3x+5}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=\frac{x+2}{3x+5}.\text{ Then }f(x+h)=\frac{x+h+2}{3x+3h+5}.$
$\displaystyle \therefore \frac{d}{dx}(f(x))=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{\frac{x+h+2}{3x+3h+5}-\frac{x+2}{3x+5}}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{(x+h+2)(3x+5)-(x+2)(3x+3h+5)}{h(3x+3h+5)(3x+5)}$
$\displaystyle =\lim\limits_{h\to0}\frac{3x^2+3xh+11x+5h+10-(3x^2+3xh+11x+6h+10)}{h(3x+3h+5)(3x+5)}$
$\displaystyle =\lim\limits_{h\to0}\frac{-h}{h(3x+3h+5)(3x+5)}$
$\displaystyle =\lim\limits_{h\to0}\frac{-1}{(3x+3h+5)(3x+5)}$
$\displaystyle =-\frac{1}{(3x+5)^2}$
$\displaystyle \text{Hence, }\frac{d}{dx}\left(\frac{x+2}{3x+5}\right)=-\frac{1}{(3x+5)^2}.$

$\displaystyle \text{(viii) }kx^n$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=kx^n.\text{ Then }f(x+h)=k(x+h)^n.$
$\displaystyle \therefore \frac{d}{dx}(f(x))=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{k(x+h)^n-kx^n}{h}$
$\displaystyle =k\lim\limits_{h\to0}\frac{(x+h)^n-x^n}{(x+h)-x}$
$\displaystyle \text{Let }z=x+h.\text{ Then }z\to x\text{ as }h\to0.$
$\displaystyle \therefore \frac{d}{dx}(f(x))=k\lim\limits_{z\to x}\frac{z^n-x^n}{z-x}$
$\displaystyle =k\left(nx^{n-1}\right)=knx^{n-1}$
$\displaystyle \text{Hence, }\frac{d}{dx}(kx^n)=knx^{n-1}.$

$\displaystyle \text{(ix) }\frac{1}{\sqrt{3-x}}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=\frac{1}{\sqrt{3-x}}.\text{ Then }f(x+h)=\frac{1}{\sqrt{3-x-h}}.$
$\displaystyle \therefore \frac{d}{dx}(f(x))=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{\frac{1}{\sqrt{3-x-h}}-\frac{1}{\sqrt{3-x}}}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{\sqrt{3-x}-\sqrt{3-x-h}}{h\sqrt{3-x}\sqrt{3-x-h}}$
$\displaystyle =\lim\limits_{h\to0}\frac{\sqrt{3-x}-\sqrt{3-x-h}}{h\sqrt{3-x}\sqrt{3-x-h}}\times\frac{\sqrt{3-x}+\sqrt{3-x-h}}{\sqrt{3-x}+\sqrt{3-x-h}}$
$\displaystyle =\lim\limits_{h\to0}\frac{(3-x)-(3-x-h)}{h\sqrt{3-x}\sqrt{3-x-h}\left(\sqrt{3-x}+\sqrt{3-x-h}\right)}$
$\displaystyle =\lim\limits_{h\to0}\frac{h}{h\sqrt{3-x}\sqrt{3-x-h}\left(\sqrt{3-x}+\sqrt{3-x-h}\right)}$
$\displaystyle =\lim\limits_{h\to0}\frac{1}{\sqrt{3-x}\sqrt{3-x-h}\left(\sqrt{3-x}+\sqrt{3-x-h}\right)}$
$\displaystyle =\frac{1}{\sqrt{3-x}\sqrt{3-x}\left(\sqrt{3-x}+\sqrt{3-x}\right)}$
$\displaystyle =\frac{1}{2(3-x)\sqrt{3-x}}$
$\displaystyle =\frac{1}{2(3-x)^{\frac{3}{2}}}$
$\displaystyle \text{Hence, }\frac{d}{dx}\left(\frac{1}{\sqrt{3-x}}\right)=\frac{1}{2(3-x)^{\frac{3}{2}}}.$

$\displaystyle \text{(x) }x^2+x+3$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=x^2+x+3.\text{ Then }f(x+h)=(x+h)^2+x+h+3.$
$\displaystyle \therefore \frac{d}{dx}(f(x))=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{(x+h)^2+x+h+3-(x^2+x+3)}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{x^2+2xh+h^2+x+h+3-x^2-x-3}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{h^2+2xh+h}{h}$
$\displaystyle =\lim\limits_{h\to0}(h+2x+1)$
$\displaystyle =2x+1$
$\displaystyle \text{Hence, }\frac{d}{dx}(x^2+x+3)=2x+1.$

$\displaystyle \text{(xi) }(x+2)^3$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=(x+2)^3.\text{ Then }f(x+h)=(x+h+2)^3.$
$\displaystyle \therefore \frac{d}{dx}(f(x))=\lim\limits_{h\to0}\frac{(x+h+2)^3-(x+2)^3}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{(x+2)^3+3(x+2)^2h+3(x+2)h^2+h^3-(x+2)^3}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{3(x+2)^2h+3(x+2)h^2+h^3}{h}$
$\displaystyle =\lim\limits_{h\to0}\left[3(x+2)^2+3(x+2)h+h^2\right]$
$\displaystyle =3(x+2)^2$
$\displaystyle \text{Hence, }\frac{d}{dx}\left((x+2)^3\right)=3(x+2)^2.$

$\displaystyle \text{(xii) }x^3+4x^2+3x+2$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=x^3+4x^2+3x+2.$
$\displaystyle \therefore \frac{d}{dx}(f(x))=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{(x+h)^3+4(x+h)^2+3(x+h)+2-(x^3+4x^2+3x+2)}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{3hx^2+3h^2x+h^3+8hx+4h^2+3h}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{h(3x^2+3hx+h^2+8x+4h+3)}{h}$
$\displaystyle =\lim\limits_{h\to0}(3x^2+3hx+h^2+8x+4h+3)$
$\displaystyle =3x^2+8x+3$
$\displaystyle \text{Hence, }\frac{d}{dx}(x^3+4x^2+3x+2)=3x^2+8x+3.$

$\displaystyle \text{(xiii) }(x^2+1)(x-5)$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=(x^2+1)(x-5).$
$\displaystyle \text{Then }f(x+h)=\left((x+h)^2+1\right)(x+h-5).$
$\displaystyle \therefore \frac{d}{dx}(f(x))=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{\left((x+h)^2+1\right)(x+h-5)-(x^2+1)(x-5)}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{3x^2h+3xh^2-10xh+h^3-5h^2+h}{h}$
$\displaystyle =\lim\limits_{h\to0}\left(3x^2+3xh-10x+h^2-5h+1\right)$
$\displaystyle =3x^2-10x+1$
$\displaystyle \text{Hence, }\frac{d}{dx}\left((x^2+1)(x-5)\right)=3x^2-10x+1.$

$\displaystyle \text{(xiv) }\sqrt{2x^2+1}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=\sqrt{2x^2+1}.$
$\displaystyle \text{Then }f(x+h)=\sqrt{2(x+h)^2+1}.$
$\displaystyle \therefore \frac{d}{dx}(f(x))=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{\sqrt{2(x+h)^2+1}-\sqrt{2x^2+1}}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{\sqrt{2x^2+4xh+2h^2+1}-\sqrt{2x^2+1}}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{\sqrt{2x^2+4xh+2h^2+1}-\sqrt{2x^2+1}}{h}$
$\displaystyle \qquad\times\frac{\sqrt{2x^2+4xh+2h^2+1}+\sqrt{2x^2+1}}{\sqrt{2x^2+4xh+2h^2+1}+\sqrt{2x^2+1}}$
$\displaystyle =\lim\limits_{h\to0}\frac{4xh+2h^2}{h\left(\sqrt{2x^2+4xh+2h^2+1}+\sqrt{2x^2+1}\right)}$
$\displaystyle =\lim\limits_{h\to0}\frac{4x+2h}{\sqrt{2x^2+4xh+2h^2+1}+\sqrt{2x^2+1}}$
$\displaystyle =\frac{4x}{2\sqrt{2x^2+1}}$
$\displaystyle =\frac{2x}{\sqrt{2x^2+1}}$
$\displaystyle \text{Hence, }\frac{d}{dx}\left(\sqrt{2x^2+1}\right)=\frac{2x}{\sqrt{2x^2+1}}.$

$\displaystyle \text{(xv) }\frac{2x+3}{x-2}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=\frac{2x+3}{x-2}.$
$\displaystyle \text{Then }f(x+h)=\frac{2x+2h+3}{x+h-2}.$
$\displaystyle \therefore \frac{d}{dx}(f(x))=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{\frac{2x+2h+3}{x+h-2}-\frac{2x+3}{x-2}}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{(2x+2h+3)(x-2)-(2x+3)(x+h-2)}{h(x+h-2)(x-2)}$
$\displaystyle =\lim\limits_{h\to0}\frac{2x^2+2xh-x-4h-6-(2x^2+2xh-x+3h-6)}{h(x+h-2)(x-2)}$
$\displaystyle =\lim\limits_{h\to0}\frac{-7h}{h(x+h-2)(x-2)}$
$\displaystyle =\lim\limits_{h\to0}\frac{-7}{(x+h-2)(x-2)}$
$\displaystyle =-\frac{7}{(x-2)^2}$
$\displaystyle \text{Hence, }\frac{d}{dx}\left(\frac{2x+3}{x-2}\right)=-\frac{7}{(x-2)^2}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: } \text{Differentiate each of the following from first principles:}$
$\displaystyle \text{(i) }e^{-x}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=e^{-x}.\text{ Then }f(x+h)=e^{-(x+h)}.$
$\displaystyle \frac{d}{dx}(f(x))=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{e^{-(x+h)}-e^{-x}}{h}$
$\displaystyle =e^{-x}\lim\limits_{h\to0}\frac{e^{-h}-1}{h}$
$\displaystyle =-e^{-x}\lim\limits_{h\to0}\frac{e^{-h}-1}{-h}$
$\displaystyle =-e^{-x}$
$\displaystyle \text{Hence, }\frac{d}{dx}(e^{-x})=-e^{-x}.$

$\displaystyle \text{(ii) }e^{3x}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=e^{3x}.\text{ Then }f(x+h)=e^{3(x+h)}.$
$\displaystyle \frac{d}{dx}(f(x))=\lim\limits_{h\to0}\frac{e^{3(x+h)}-e^{3x}}{h}$
$\displaystyle =e^{3x}\lim\limits_{h\to0}\frac{e^{3h}-1}{h}$
$\displaystyle =3e^{3x}\lim\limits_{h\to0}\frac{e^{3h}-1}{3h}$
$\displaystyle =3e^{3x}$
$\displaystyle \text{Hence, }\frac{d}{dx}(e^{3x})=3e^{3x}.$

$\displaystyle \text{(iii) }e^{ax+b}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=e^{ax+b}.\text{ Then }f(x+h)=e^{a(x+h)+b}.$
$\displaystyle \frac{d}{dx}(f(x))=\lim\limits_{h\to0}\frac{e^{a(x+h)+b}-e^{ax+b}}{h}$
$\displaystyle =e^{ax+b}\lim\limits_{h\to0}\frac{e^{ah}-1}{h}$
$\displaystyle =ae^{ax+b}\lim\limits_{h\to0}\frac{e^{ah}-1}{ah}$
$\displaystyle =ae^{ax+b}$
$\displaystyle \text{Hence, }\frac{d}{dx}(e^{ax+b})=ae^{ax+b}.$

$\displaystyle \text{(iv) }xe^x$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=xe^x.\text{ Then }f(x+h)=(x+h)e^{x+h}.$
$\displaystyle \frac{d}{dx}(f(x))=\lim\limits_{h\to0}\frac{(x+h)e^{x+h}-xe^x}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{(x+h)e^xe^h-xe^x}{h}$
$\displaystyle =xe^x\lim\limits_{h\to0}\frac{e^h-1}{h} +e^x\lim\limits_{h\to0}e^h$
$\displaystyle =xe^x+e^x$
$\displaystyle =e^x(x+1)$
$\displaystyle \text{Hence, }\frac{d}{dx}(xe^x)=e^x(x+1).$

$\displaystyle \text{(v) }-x$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(v) Let }f(x)=-x.\text{ Then }f(x+h)=-(x+h).$
$\displaystyle \frac{d}{dx}(f(x))=\lim\limits_{h\to0}\frac{-(x+h)-(-x)}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{-h}{h}$
$\displaystyle =-1$
$\displaystyle \text{Hence, }\frac{d}{dx}(-x)=-1.$

$\displaystyle \text{(vi) }(-x)^{-1}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=(-x)^{-1}=-\frac{1}{x}.$
$\displaystyle \text{Then }f(x+h)=-\frac{1}{x+h}.$
$\displaystyle \frac{d}{dx}(f(x))=\lim\limits_{h\to0} \frac{-\frac{1}{x+h}+\frac{1}{x}}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{h}{hx(x+h)}$
$\displaystyle =\lim\limits_{h\to0}\frac{1}{x(x+h)}$
$\displaystyle =\frac{1}{x^2}$
$\displaystyle \text{Hence, }\frac{d}{dx}\left((-x)^{-1}\right)=\frac{1}{x^2}.$

$\displaystyle \text{(vii) }\sin(x+1)$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=\sin(x+1).$
$\displaystyle \text{Then }f(x+h)=\sin(x+h+1).$
$\displaystyle \frac{d}{dx}(f(x))=\lim\limits_{h\to0} \frac{\sin(x+h+1)-\sin(x+1)}{h}$
$\displaystyle =\lim\limits_{h\to0} \frac{2\cos\left(x+1+\frac{h}{2}\right)\sin\frac{h}{2}}{h}$
$\displaystyle =\lim\limits_{h\to0}\cos\left(x+1+\frac{h}{2}\right) \frac{\sin\frac{h}{2}}{\frac{h}{2}}$
$\displaystyle =\cos(x+1)$
$\displaystyle \text{Hence, }\frac{d}{dx}\left(\sin(x+1)\right)=\cos(x+1).$

$\displaystyle \text{(viii) }\cos\left(x-\frac{\pi}{8}\right)$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=\cos\left(x-\frac{\pi}{8}\right).$
$\displaystyle \text{Then }f(x+h)=\cos\left(x+h-\frac{\pi}{8}\right).$
$\displaystyle \frac{d}{dx}(f(x))=\lim\limits_{h\to0} \frac{\cos\left(x+h-\frac{\pi}{8}\right)-\cos\left(x-\frac{\pi}{8}\right)}{h}$
$\displaystyle =\lim\limits_{h\to0} \frac{-2\sin\left(x-\frac{\pi}{8}+\frac{h}{2}\right)\sin\frac{h}{2}}{h}$
$\displaystyle =-\lim\limits_{h\to0} \sin\left(x-\frac{\pi}{8}+\frac{h}{2}\right) \frac{\sin\frac{h}{2}}{\frac{h}{2}}$
$\displaystyle =-\sin\left(x-\frac{\pi}{8}\right)$
$\displaystyle \text{Hence, }\frac{d}{dx}\left(\cos\left(x-\frac{\pi}{8}\right)\right) =-\sin\left(x-\frac{\pi}{8}\right).$

$\displaystyle \text{(ix) }x\sin x$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=x\sin x.$
$\displaystyle \text{Then }f(x+h)=(x+h)\sin(x+h).$
$\displaystyle \frac{d}{dx}(f(x))=\lim\limits_{h\to0} \frac{(x+h)\sin(x+h)-x\sin x}{h}$
$\displaystyle =\lim\limits_{h\to0} \frac{(x+h)(\sin x\cos h+\cos x\sin h)-x\sin x}{h}$
$\displaystyle =x\sin x\lim\limits_{h\to0}\frac{\cos h-1}{h} +x\cos x\lim\limits_{h\to0}\frac{\sin h}{h}$
$\displaystyle \qquad+\sin x\lim\limits_{h\to0}\cos h +\cos x\lim\limits_{h\to0}\sin h$
$\displaystyle \text{Now, }\lim\limits_{h\to0}\frac{\cos h-1}{h} =-\lim\limits_{h\to0}\frac{\sin\frac{h}{2}}{\frac{h}{2}}\sin\frac{h}{2}=0.$
$\displaystyle \therefore \frac{d}{dx}(f(x)) =x\sin x(0)+x\cos x(1)+\sin x(1)+\cos x(0)$
$\displaystyle =x\cos x+\sin x$
$\displaystyle \text{Hence, }\frac{d}{dx}(x\sin x)=x\cos x+\sin x.$

$\displaystyle \text{(x) }x\cos x$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=x\cos x.$
$\displaystyle \text{Then }f(x+h)=(x+h)\cos(x+h).$
$\displaystyle \frac{d}{dx}(f(x))=\lim\limits_{h\to0} \frac{(x+h)\cos(x+h)-x\cos x}{h}$
$\displaystyle =\lim\limits_{h\to0} \frac{(x+h)(\cos x\cos h-\sin x\sin h)-x\cos x}{h}$
$\displaystyle =x\cos x\lim\limits_{h\to0}\frac{\cos h-1}{h} -x\sin x\lim\limits_{h\to0}\frac{\sin h}{h}$
$\displaystyle \qquad+\cos x\lim\limits_{h\to0}\cos h -\sin x\lim\limits_{h\to0}\sin h$
$\displaystyle =x\cos x(0)-x\sin x(1)+\cos x(1)-\sin x(0)$
$\displaystyle =\cos x-x\sin x$
$\displaystyle \text{Hence, }\frac{d}{dx}(x\cos x)=\cos x-x\sin x.$

$\displaystyle \text{(xi) }\sin(2x-3)$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=\sin(2x-3).$
$\displaystyle \text{Then }f(x+h)=\sin(2x+2h-3).$
$\displaystyle \frac{d}{dx}(f(x))=\lim\limits_{h\to0} \frac{\sin(2x+2h-3)-\sin(2x-3)}{h}$
$\displaystyle =\lim\limits_{h\to0} \frac{2\cos\left(\frac{(2x+2h-3)+(2x-3)}{2}\right) \sin\left(\frac{(2x+2h-3)-(2x-3)}{2}\right)}{h}$
$\displaystyle =\lim\limits_{h\to0} 2\cos(2x+h-3)\frac{\sin h}{h}$
$\displaystyle =2\cos(2x-3)$
$\displaystyle \text{Hence, }\frac{d}{dx}\left(\sin(2x-3)\right)=2\cos(2x-3).$

$\displaystyle \textbf{Question 3: } \text{Differentiate the following from first principles:}$
$\displaystyle \text{(i) }\sqrt{\sin 2x}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Let }f(x)=\sqrt{\sin 2x}.$
$\displaystyle \text{Then }f(x+h)=\sqrt{\sin(2x+2h)}.$
$\displaystyle \frac{d}{dx}(f(x))=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{\sqrt{\sin(2x+2h)}-\sqrt{\sin 2x}}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{\sqrt{\sin(2x+2h)}-\sqrt{\sin 2x}}{h}$
$\displaystyle \qquad\times\frac{\sqrt{\sin(2x+2h)}+\sqrt{\sin 2x}}{\sqrt{\sin(2x+2h)}+\sqrt{\sin 2x}}$
$\displaystyle =\lim\limits_{h\to0}\frac{\sin(2x+2h)-\sin 2x}{h\left(\sqrt{\sin(2x+2h)}+\sqrt{\sin 2x}\right)}$
$\displaystyle =\lim\limits_{h\to0}\frac{2\cos\left(\frac{2x+2h+2x}{2}\right)\sin\left(\frac{2x+2h-2x}{2}\right)}{h\left(\sqrt{\sin(2x+2h)}+\sqrt{\sin 2x}\right)}$
$\displaystyle =\lim\limits_{h\to0}\frac{2\cos(2x+h)\sin h}{h\left(\sqrt{\sin(2x+2h)}+\sqrt{\sin 2x}\right)}$
$\displaystyle =\lim\limits_{h\to0}2\cos(2x+h)\frac{\sin h}{h}\frac{1}{\sqrt{\sin(2x+2h)}+\sqrt{\sin 2x}}$
$\displaystyle =2\cos 2x\cdot1\cdot\frac{1}{2\sqrt{\sin 2x}}$
$\displaystyle =\frac{\cos 2x}{\sqrt{\sin 2x}}$
$\displaystyle \text{Hence, }\frac{d}{dx}\left(\sqrt{\sin 2x}\right)=\frac{\cos 2x}{\sqrt{\sin 2x}}.$

$\displaystyle \text{(ii) }\frac{\sin x}{x}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=\frac{\sin x}{x}.\text{ Then }f(x+h)=\frac{\sin(x+h)}{x+h}.$
$\displaystyle \frac{d}{dx}(f(x))=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{\frac{\sin(x+h)}{x+h}-\frac{\sin x}{x}}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{x\sin(x+h)-(x+h)\sin x}{hx(x+h)}$
$\displaystyle =\lim\limits_{h\to0}\frac{x(\sin x\cos h+\cos x\sin h)-x\sin x-h\sin x}{hx(x+h)}$
$\displaystyle =\lim\limits_{h\to0}\frac{x\sin x(\cos h-1)+x\cos x\sin h-h\sin x}{hx(x+h)}$
$\displaystyle =\sin x\lim\limits_{h\to0}\frac{\cos h-1}{h}\frac{1}{x+h}$
$\displaystyle \qquad+\cos x\lim\limits_{h\to0}\frac{\sin h}{h}\frac{1}{x+h}$
$\displaystyle \qquad-\frac{\sin x}{x}\lim\limits_{h\to0}\frac{1}{x+h}$
$\displaystyle \text{Now, }\lim\limits_{h\to0}\frac{\cos h-1}{h}$
$\displaystyle =-\lim\limits_{h\to0}\frac{\sin\frac{h}{2}}{\frac{h}{2}}\sin\frac{h}{2}=0.$
$\displaystyle \therefore \frac{d}{dx}(f(x))$
$\displaystyle =\sin x(0)\left(\frac{1}{x}\right)+\cos x(1)\left(\frac{1}{x}\right)-\frac{\sin x}{x}\left(\frac{1}{x}\right)$
$\displaystyle =\frac{\cos x}{x}-\frac{\sin x}{x^2}$
$\displaystyle =\frac{x\cos x-\sin x}{x^2}$
$\displaystyle \text{Hence, }\frac{d}{dx}\left(\frac{\sin x}{x}\right)=\frac{x\cos x-\sin x}{x^2},\quad x\ne0.$

$\displaystyle \text{(iii) }\frac{\cos x}{x}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=\frac{\cos x}{x}.\text{ Then }f(x+h)=\frac{\cos(x+h)}{x+h}.$
$\displaystyle \frac{d}{dx}(f(x))=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{\frac{\cos(x+h)}{x+h}-\frac{\cos x}{x}}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{x\cos(x+h)-(x+h)\cos x}{hx(x+h)}$
$\displaystyle =\lim\limits_{h\to0}\frac{x(\cos x\cos h-\sin x\sin h)-x\cos x-h\cos x}{hx(x+h)}$
$\displaystyle =\lim\limits_{h\to0}\frac{x\cos x(\cos h-1)-x\sin x\sin h-h\cos x}{hx(x+h)}$
$\displaystyle =\cos x\lim\limits_{h\to0}\frac{\cos h-1}{h}\frac{1}{x+h}$
$\displaystyle \qquad-\sin x\lim\limits_{h\to0}\frac{\sin h}{h}\frac{1}{x+h}$
$\displaystyle \qquad-\frac{\cos x}{x}\lim\limits_{h\to0}\frac{1}{x+h}$
$\displaystyle \text{Now, }\lim\limits_{h\to0}\frac{\cos h-1}{h}$
$\displaystyle =-\lim\limits_{h\to0}\frac{\sin\frac{h}{2}}{\frac{h}{2}}\sin\frac{h}{2}=0.$
$\displaystyle \therefore \frac{d}{dx}(f(x))$
$\displaystyle =\cos x(0)\left(\frac{1}{x}\right)-\sin x(1)\left(\frac{1}{x}\right)-\frac{\cos x}{x}\left(\frac{1}{x}\right)$
$\displaystyle =-\frac{\sin x}{x}-\frac{\cos x}{x^2}$
$\displaystyle =\frac{-x\sin x-\cos x}{x^2}$
$\displaystyle \text{Hence, }\frac{d}{dx}\left(\frac{\cos x}{x}\right)=\frac{-x\sin x-\cos x}{x^2},\quad x\ne0.$

$\displaystyle \text{(iv) }x^2\sin x$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=x^2\sin x.\text{ Then }f(x+h)=(x+h)^2\sin(x+h).$
$\displaystyle \frac{d}{dx}(f(x))=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{(x+h)^2\sin(x+h)-x^2\sin x}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{(x^2+2xh+h^2)(\sin x\cos h+\cos x\sin h)-x^2\sin x}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{x^2\sin x(\cos h-1)+x^2\cos x\sin h+h^2\sin x\cos h+h^2\cos x\sin h+2xh\sin x\cos h+2xh\cos x\sin h}{h}$
$\displaystyle =x^2\sin x\lim\limits_{h\to0}\frac{\cos h-1}{h}+x^2\cos x\lim\limits_{h\to0}\frac{\sin h}{h}$
$\displaystyle \qquad+\sin x\lim\limits_{h\to0}h\cos h+\cos x\lim\limits_{h\to0}h\sin h$
$\displaystyle \qquad+2x\sin x\lim\limits_{h\to0}\cos h+2x\cos x\lim\limits_{h\to0}\sin h$
$\displaystyle \text{Now, }\lim\limits_{h\to0}\frac{\cos h-1}{h}=-\lim\limits_{h\to0}\frac{\sin\frac{h}{2}}{\frac{h}{2}}\sin\frac{h}{2}=0.$
$\displaystyle \therefore \frac{d}{dx}(f(x))=x^2\sin x(0)+x^2\cos x(1)+\sin x(0)+\cos x(0)+2x\sin x(1)+2x\cos x(0)$
$\displaystyle =x^2\cos x+2x\sin x$
$\displaystyle \text{Hence, }\frac{d}{dx}(x^2\sin x)=x^2\cos x+2x\sin x.$

$\displaystyle \text{(v) }\sqrt{\sin(3x+1)}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=\sqrt{\sin(3x+1)}.$
$\displaystyle \text{Then }f(x+h)=\sqrt{\sin(3x+3h+1)}.$
$\displaystyle \frac{d}{dx}(f(x))=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{\sqrt{\sin(3x+3h+1)}-\sqrt{\sin(3x+1)}}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{\sqrt{\sin(3x+3h+1)}-\sqrt{\sin(3x+1)}}{h}$
$\displaystyle \qquad\times\frac{\sqrt{\sin(3x+3h+1)}+\sqrt{\sin(3x+1)}}{\sqrt{\sin(3x+3h+1)}+\sqrt{\sin(3x+1)}}$
$\displaystyle =\lim\limits_{h\to0}\frac{\sin(3x+3h+1)-\sin(3x+1)}{h\left(\sqrt{\sin(3x+3h+1)}+\sqrt{\sin(3x+1)}\right)}$
$\displaystyle =\lim\limits_{h\to0}\frac{2\cos\left(\frac{(3x+3h+1)+(3x+1)}{2}\right)\sin\left(\frac{(3x+3h+1)-(3x+1)}{2}\right)}{h\left(\sqrt{\sin(3x+3h+1)}+\sqrt{\sin(3x+1)}\right)}$
$\displaystyle =\lim\limits_{h\to0}\frac{2\cos\left(3x+\frac{3h}{2}+1\right)\sin\frac{3h}{2}}{h\left(\sqrt{\sin(3x+3h+1)}+\sqrt{\sin(3x+1)}\right)}$
$\displaystyle =3\lim\limits_{h\to0}\cos\left(3x+\frac{3h}{2}+1\right)\frac{\sin\frac{3h}{2}}{\frac{3h}{2}}$
$\displaystyle \qquad\times\frac{1}{\sqrt{\sin(3x+3h+1)}+\sqrt{\sin(3x+1)}}$
$\displaystyle =3\cos(3x+1)\cdot1\cdot\frac{1}{2\sqrt{\sin(3x+1)}}$
$\displaystyle =\frac{3\cos(3x+1)}{2\sqrt{\sin(3x+1)}}$
$\displaystyle \text{Hence, }\frac{d}{dx}\left(\sqrt{\sin(3x+1)}\right)=\frac{3\cos(3x+1)}{2\sqrt{\sin(3x+1)}}.$

$\displaystyle \text{(vi) }\sin x+\cos x$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=\sin x+\cos x.$
$\displaystyle \text{Then }f(x+h)=\sin(x+h)+\cos(x+h).$
$\displaystyle \frac{d}{dx}(f(x))=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{\sin(x+h)+\cos(x+h)-\sin x-\cos x}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{\sin(x+h)-\sin x}{h}+\lim\limits_{h\to0}\frac{\cos(x+h)-\cos x}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{2\cos\left(x+\frac h2\right)\sin\frac h2}{h}$
$\displaystyle \qquad-\lim\limits_{h\to0}\frac{2\sin\left(x+\frac h2\right)\sin\frac h2}{h}$
$\displaystyle =\lim\limits_{h\to0}\cos\left(x+\frac h2\right)\frac{\sin\frac h2}{\frac h2}$
$\displaystyle \qquad-\lim\limits_{h\to0}\sin\left(x+\frac h2\right)\frac{\sin\frac h2}{\frac h2}$
$\displaystyle =\cos x-\sin x$
$\displaystyle \text{Hence, }\frac{d}{dx}(\sin x+\cos x)=\cos x-\sin x.$

$\displaystyle \text{(vii) }x^2e^x$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=x^2e^x.$
$\displaystyle \text{Then }f(x+h)=(x+h)^2e^{x+h}.$
$\displaystyle \frac{d}{dx}(f(x))=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{(x+h)^2e^{x+h}-x^2e^x}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{(x^2+2xh+h^2)e^xe^h-x^2e^x}{h}$
$\displaystyle =\lim\limits_{h\to0}\left[x^2e^x\frac{e^h-1}{h}+2xe^xe^h+he^xe^h\right]$
$\displaystyle =x^2e^x(1)+2xe^x(1)+0$
$\displaystyle =x^2e^x+2xe^x$
$\displaystyle =(x^2+2x)e^x$
$\displaystyle \text{Hence, }\frac{d}{dx}(x^2e^x)=(x^2+2x)e^x.$

$\displaystyle \text{(viii) }e^{x^2+1}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=e^{x^2+1}.$
$\displaystyle \text{Then }f(x+h)=e^{(x+h)^2+1}.$
$\displaystyle \frac{d}{dx}(f(x))=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{e^{(x+h)^2+1}-e^{x^2+1}}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{e^{x^2+2xh+h^2+1}-e^{x^2+1}}{h}$
$\displaystyle =e^{x^2+1}\lim\limits_{h\to0}\frac{e^{h(2x+h)}-1}{h}$
$\displaystyle =e^{x^2+1}\lim\limits_{h\to0}\frac{e^{h(2x+h)}-1}{h(2x+h)}(2x+h)$
$\displaystyle =e^{x^2+1}(1)(2x)$
$\displaystyle =2xe^{x^2+1}$
$\displaystyle \text{Hence, }\frac{d}{dx}\left(e^{x^2+1}\right)=2xe^{x^2+1}.$

$\displaystyle \text{(ix) }e^{\sqrt{2x}}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=e^{\sqrt{2x}}.$
$\displaystyle \text{Then }f(x+h)=e^{\sqrt{2x+2h}}.$
$\displaystyle \frac{d}{dx}(f(x))=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{e^{\sqrt{2x+2h}}-e^{\sqrt{2x}}}{h}$
$\displaystyle =2\lim\limits_{h\to0}\frac{e^{\sqrt{2x+2h}}-e^{\sqrt{2x}}}{(2x+2h)-2x}$
$\displaystyle =2\lim\limits_{h\to0}\frac{e^{\sqrt{2x+2h}}-e^{\sqrt{2x}}}{(\sqrt{2x+2h})^2-(\sqrt{2x})^2}$
$\displaystyle =2e^{\sqrt{2x}}\lim\limits_{h\to0}\frac{e^{\sqrt{2x+2h}-\sqrt{2x}}-1}{\sqrt{2x+2h}-\sqrt{2x}}$
$\displaystyle \qquad\times\lim\limits_{h\to0}\frac{1}{\sqrt{2x+2h}+\sqrt{2x}}$
$\displaystyle =2e^{\sqrt{2x}}\cdot1\cdot\frac{1}{2\sqrt{2x}}$
$\displaystyle =\frac{e^{\sqrt{2x}}}{\sqrt{2x}}$
$\displaystyle \text{Hence, }\frac{d}{dx}\left(e^{\sqrt{2x}}\right)=\frac{e^{\sqrt{2x}}}{\sqrt{2x}},\quad x>0.$
$\displaystyle \\$

$\displaystyle \text{(x) }e^{\sqrt{ax+b}}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=e^{\sqrt{ax+b}}.$
$\displaystyle \text{Then }f(x+h)=e^{\sqrt{ax+ah+b}}.$
$\displaystyle \frac{d}{dx}(f(x))=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{e^{\sqrt{ax+ah+b}}-e^{\sqrt{ax+b}}}{h}$
$\displaystyle =a\lim\limits_{h\to0}\frac{e^{\sqrt{ax+ah+b}}-e^{\sqrt{ax+b}}}{(ax+ah+b)-(ax+b)}$
$\displaystyle =ae^{\sqrt{ax+b}}\lim\limits_{h\to0}\frac{e^{\sqrt{ax+ah+b}-\sqrt{ax+b}}-1}{(\sqrt{ax+ah+b})^2-(\sqrt{ax+b})^2}$
$\displaystyle =ae^{\sqrt{ax+b}}\lim\limits_{h\to0}\frac{e^{\sqrt{ax+ah+b}-\sqrt{ax+b}}-1}{\sqrt{ax+ah+b}-\sqrt{ax+b}}$
$\displaystyle \qquad\times\lim\limits_{h\to0}\frac{1}{\sqrt{ax+ah+b}+\sqrt{ax+b}}$
$\displaystyle =ae^{\sqrt{ax+b}}\cdot1\cdot\frac{1}{2\sqrt{ax+b}}$
$\displaystyle =\frac{ae^{\sqrt{ax+b}}}{2\sqrt{ax+b}}$
$\displaystyle \text{Hence, }\frac{d}{dx}\left(e^{\sqrt{ax+b}}\right)=\frac{ae^{\sqrt{ax+b}}}{2\sqrt{ax+b}}.$

$\displaystyle \text{(xi) }a^x$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=a^x.$
$\displaystyle \text{Then }f(x+h)=a^{x+h}.$
$\displaystyle \frac{d}{dx}(f(x))=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{a^{x+h}-a^x}{h}$
$\displaystyle =a^x\lim\limits_{h\to0}\frac{a^h-1}{h}$
$\displaystyle =a^x\log_e a$
$\displaystyle \text{Hence, }\frac{d}{dx}(a^x)=a^x\log_e a,\quad a>0.$
$\displaystyle \\$

$\displaystyle \text{(xii) }3^{x^2}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=3^{x^2}.$
$\displaystyle \text{Then }f(x+h)=3^{(x+h)^2}.$
$\displaystyle \frac{d}{dx}(f(x))=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{3^{(x+h)^2}-3^{x^2}}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{3^{x^2+2xh+h^2}-3^{x^2}}{h}$
$\displaystyle =3^{x^2}\lim\limits_{h\to0}\frac{3^{h(2x+h)}-1}{h}$
$\displaystyle =3^{x^2}\lim\limits_{h\to0}\frac{3^{h(2x+h)}-1}{h(2x+h)}(2x+h)$
$\displaystyle =3^{x^2}\log_e3\,(2x)$
$\displaystyle =2x\,3^{x^2}\log_e3$
$\displaystyle \text{Hence, }\frac{d}{dx}\left(3^{x^2}\right)=2x\,3^{x^2}\log_e3.$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{Differentiate each of the following from first principles:}$
$\displaystyle \text{(i) }\tan^2x \qquad \text{(ii) }\tan(2x+1) \qquad \text{(iii) }\tan 2x \qquad \text{(iv) }\sqrt{\tan x}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Let }f(x)=\tan^2x.$
$\displaystyle \text{Then }f(x+h)=\tan^2(x+h).$
$\displaystyle \frac{d}{dx}(f(x))=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{\tan^2(x+h)-\tan^2x}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{[\tan(x+h)-\tan x][\tan(x+h)+\tan x]}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{[\sin(x+h)\cos x-\cos(x+h)\sin x]}{h\cos(x+h)\cos x}$
$\displaystyle \qquad\times\frac{[\sin(x+h)\cos x+\cos(x+h)\sin x]}{\cos(x+h)\cos x}$
$\displaystyle =\lim\limits_{h\to0}\frac{\sin h\sin(2x+h)}{h\cos^2(x+h)\cos^2x}$
$\displaystyle =\frac{1}{\cos^2x}\lim\limits_{h\to0}\frac{\sin h}{h}\lim\limits_{h\to0}\sin(2x+h)\lim\limits_{h\to0}\frac{1}{\cos^2(x+h)}$
$\displaystyle =\frac{1}{\cos^2x}\cdot1\cdot\sin2x\cdot\frac{1}{\cos^2x}$
$\displaystyle =\frac{2\sin x\cos x}{\cos^4x}$
$\displaystyle =2\tan x\sec^2x$
$\displaystyle \text{Hence, }\frac{d}{dx}(\tan^2x)=2\tan x\sec^2x.$

$\displaystyle \text{(ii) Let }f(x)=\tan(2x+1).$
$\displaystyle \text{Then }f(x+h)=\tan(2x+2h+1).$
$\displaystyle \frac{d}{dx}(f(x))=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{\tan(2x+2h+1)-\tan(2x+1)}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{\sin(2x+2h+1)\cos(2x+1)-\cos(2x+2h+1)\sin(2x+1)}{h\cos(2x+2h+1)\cos(2x+1)}$
$\displaystyle =\lim\limits_{h\to0}\frac{\sin2h}{h\cos(2x+2h+1)\cos(2x+1)}$
$\displaystyle =2\lim\limits_{h\to0}\frac{\sin2h}{2h}\lim\limits_{h\to0}\frac{1}{\cos(2x+2h+1)\cos(2x+1)}$
$\displaystyle =2\cdot1\cdot\frac{1}{\cos^2(2x+1)}$
$\displaystyle =2\sec^2(2x+1)$
$\displaystyle \text{Hence, }\frac{d}{dx}\tan(2x+1)=2\sec^2(2x+1).$

$\displaystyle \text{(iii) Let }f(x)=\tan2x.$
$\displaystyle \text{Then }f(x+h)=\tan(2x+2h).$
$\displaystyle \frac{d}{dx}(f(x))=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{\tan(2x+2h)-\tan2x}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{\sin(2x+2h)\cos2x-\cos(2x+2h)\sin2x}{h\cos(2x+2h)\cos2x}$
$\displaystyle =\lim\limits_{h\to0}\frac{\sin2h}{h\cos(2x+2h)\cos2x}$
$\displaystyle =2\lim\limits_{h\to0}\frac{\sin2h}{2h}\lim\limits_{h\to0}\frac{1}{\cos(2x+2h)\cos2x}$
$\displaystyle =2\cdot1\cdot\frac{1}{\cos^22x}$
$\displaystyle =2\sec^22x$
$\displaystyle \text{Hence, }\frac{d}{dx}(\tan2x)=2\sec^22x.$

$\displaystyle \text{(iv) Let }f(x)=\sqrt{\tan x}.$
$\displaystyle \text{Then }f(x+h)=\sqrt{\tan(x+h)}.$
$\displaystyle \frac{d}{dx}(f(x))=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{\sqrt{\tan(x+h)}-\sqrt{\tan x}}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{\sqrt{\tan(x+h)}-\sqrt{\tan x}}{h}\times\frac{\sqrt{\tan(x+h)}+\sqrt{\tan x}}{\sqrt{\tan(x+h)}+\sqrt{\tan x}}$
$\displaystyle =\lim\limits_{h\to0}\frac{\tan(x+h)-\tan x}{h[\sqrt{\tan(x+h)}+\sqrt{\tan x}]}$
$\displaystyle =\lim\limits_{h\to0}\frac{\sin(x+h)\cos x-\cos(x+h)\sin x}{h[\sqrt{\tan(x+h)}+\sqrt{\tan x}]\cos(x+h)\cos x}$
$\displaystyle =\lim\limits_{h\to0}\frac{\sin h}{h}\frac{1}{[\sqrt{\tan(x+h)}+\sqrt{\tan x}]\cos(x+h)\cos x}$
$\displaystyle =1\cdot\frac{1}{2\sqrt{\tan x}\cos^2x}$
$\displaystyle =\frac{\sec^2x}{2\sqrt{\tan x}}$
$\displaystyle \text{Hence, }\frac{d}{dx}\left(\sqrt{\tan x}\right)=\frac{\sec^2x}{2\sqrt{\tan x}}.$

$\displaystyle \textbf{Question 5: }\text{Differentiate each of the following from first principles:}$
$\displaystyle \text{(i) }\sin\sqrt{2x}\qquad\text{(ii) }\cos\sqrt{x}\qquad\text{(iii) }\tan\sqrt{x}\qquad\text{(iv) }\tan x^2$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Let }f(x)=\sin\sqrt{2x}.$
$\displaystyle \text{Then }f(x+h)=\sin\sqrt{2x+2h}.$
$\displaystyle \frac{d}{dx}(f(x))=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{\sin\sqrt{2x+2h}-\sin\sqrt{2x}}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{2\cos\left(\frac{\sqrt{2x+2h}+\sqrt{2x}}{2}\right)\sin\left(\frac{\sqrt{2x+2h}-\sqrt{2x}}{2}\right)}{h}$
$\displaystyle =2\lim\limits_{h\to0}\frac{2\cos\left(\frac{\sqrt{2x+2h}+\sqrt{2x}}{2}\right)\sin\left(\frac{\sqrt{2x+2h}-\sqrt{2x}}{2}\right)}{(2x+2h)-2x}$
$\displaystyle =2\lim\limits_{h\to0}\frac{2\cos\left(\frac{\sqrt{2x+2h}+\sqrt{2x}}{2}\right)\sin\left(\frac{\sqrt{2x+2h}-\sqrt{2x}}{2}\right)}{(\sqrt{2x+2h}-\sqrt{2x})(\sqrt{2x+2h}+\sqrt{2x})}$
$\displaystyle =2\lim\limits_{h\to0}\frac{\sin\left(\frac{\sqrt{2x+2h}-\sqrt{2x}}{2}\right)}{\frac{\sqrt{2x+2h}-\sqrt{2x}}{2}}$
$\displaystyle \qquad\times\lim\limits_{h\to0}\frac{\cos\left(\frac{\sqrt{2x+2h}+\sqrt{2x}}{2}\right)}{\sqrt{2x+2h}+\sqrt{2x}}$
$\displaystyle =2(1)\frac{\cos\sqrt{2x}}{2\sqrt{2x}}$
$\displaystyle =\frac{\cos\sqrt{2x}}{\sqrt{2x}}$
$\displaystyle \text{Hence, }\frac{d}{dx}\left(\sin\sqrt{2x}\right)=\frac{\cos\sqrt{2x}}{\sqrt{2x}},\quad x>0.$

$\displaystyle \text{(ii) Let }f(x)=\cos\sqrt{x}.$
$\displaystyle \text{Then }f(x+h)=\cos\sqrt{x+h}.$
$\displaystyle \frac{d}{dx}(f(x))=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{\cos\sqrt{x+h}-\cos\sqrt{x}}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{-2\sin\left(\frac{\sqrt{x+h}+\sqrt{x}}{2}\right)\sin\left(\frac{\sqrt{x+h}-\sqrt{x}}{2}\right)}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{-2\sin\left(\frac{\sqrt{x+h}+\sqrt{x}}{2}\right)\sin\left(\frac{\sqrt{x+h}-\sqrt{x}}{2}\right)}{(\sqrt{x+h}-\sqrt{x})(\sqrt{x+h}+\sqrt{x})}$
$\displaystyle =-\lim\limits_{h\to0}\sin\left(\frac{\sqrt{x+h}+\sqrt{x}}{2}\right)$
$\displaystyle \qquad\times\lim\limits_{h\to0}\frac{\sin\left(\frac{\sqrt{x+h}-\sqrt{x}}{2}\right)}{\frac{\sqrt{x+h}-\sqrt{x}}{2}}$
$\displaystyle \qquad\times\lim\limits_{h\to0}\frac{1}{\sqrt{x+h}+\sqrt{x}}$
$\displaystyle =-\sin\sqrt{x}\cdot1\cdot\frac{1}{2\sqrt{x}}$
$\displaystyle =-\frac{\sin\sqrt{x}}{2\sqrt{x}}$
$\displaystyle \text{Hence, }\frac{d}{dx}\left(\cos\sqrt{x}\right)=-\frac{\sin\sqrt{x}}{2\sqrt{x}},\quad x>0.$

$\displaystyle \text{(iii) Let }f(x)=\tan\sqrt{x}.$
$\displaystyle \text{Then }f(x+h)=\tan\sqrt{x+h}.$
$\displaystyle \frac{d}{dx}(f(x))=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{\tan\sqrt{x+h}-\tan\sqrt{x}}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{\sin(\sqrt{x+h}-\sqrt{x})}{h\cos\sqrt{x+h}\cos\sqrt{x}}$
$\displaystyle =\lim\limits_{h\to0}\frac{\sin(\sqrt{x+h}-\sqrt{x})}{(\sqrt{x+h}-\sqrt{x})(\sqrt{x+h}+\sqrt{x})\cos\sqrt{x+h}\cos\sqrt{x}}$
$\displaystyle =\lim\limits_{h\to0}\frac{\sin(\sqrt{x+h}-\sqrt{x})}{\sqrt{x+h}-\sqrt{x}}$
$\displaystyle \qquad\times\lim\limits_{h\to0}\frac{1}{(\sqrt{x+h}+\sqrt{x})\cos\sqrt{x+h}\cos\sqrt{x}}$
$\displaystyle =1\cdot\frac{1}{2\sqrt{x}\cos^2\sqrt{x}}$
$\displaystyle =\frac{\sec^2\sqrt{x}}{2\sqrt{x}}$
$\displaystyle \text{Hence, }\frac{d}{dx}\left(\tan\sqrt{x}\right)=\frac{\sec^2\sqrt{x}}{2\sqrt{x}},\quad x>0.$

$\displaystyle \text{(iv) Let }f(x)=\tan x^2.$
$\displaystyle \text{Then }f(x+h)=\tan(x+h)^2.$
$\displaystyle \frac{d}{dx}(f(x))=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{\tan(x+h)^2-\tan x^2}{h}$
$\displaystyle =\lim\limits_{h\to0}\frac{\sin\left((x+h)^2-x^2\right)}{h\cos(x+h)^2\cos x^2}$
$\displaystyle =\lim\limits_{h\to0}\frac{\sin(h^2+2xh)}{h\cos(x+h)^2\cos x^2}$
$\displaystyle =\lim\limits_{h\to0}\frac{\sin\left(h(h+2x)\right)}{h(h+2x)}$
$\displaystyle \qquad\times\lim\limits_{h\to0}\frac{h+2x}{\cos(x+h)^2\cos x^2}$
$\displaystyle =1\cdot\frac{2x}{\cos^2x^2}$
$\displaystyle =2x\sec^2x^2$
$\displaystyle \text{Hence, }\frac{d}{dx}(\tan x^2)=2x\sec^2x^2.$
$\displaystyle \\$