Class 11: Derivatives – Exercise 30.2 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Differentiate each of the following from first principle:

$\displaystyle \text{(i) } \frac{2}{x}$ $\displaystyle \text{(ii) } \frac{1}{\sqrt{x}}$ $\displaystyle \text{(iii) } \frac{1}{x^3}$

$\displaystyle \text{(vi) } \frac{x^2+1}{x}$ $\displaystyle \text{(v) } \frac{x^2-1}{x}$ $\displaystyle \text{(vi) } \frac{x+1}{x+2}$

$\displaystyle \text{(vii) } \frac{x+2}{3x+5}$ $\displaystyle \text{(viii) } kx^n$ $\displaystyle \text{(ix) } \frac{1}{\sqrt{3-x}}$

$\displaystyle \text{(x) } x^2+x+3$ $\displaystyle \text{(xi) } (x+2)^3$ $\displaystyle \text{(xii) } x^3+4x^2+3x+2$

$\displaystyle \text{(xiii) } (x^2+1)(x-5)$ $\displaystyle \text{(xiv) } \sqrt{2x^2+1}$ $\displaystyle \text{(xv) } \frac{2x+3}{x-2}$

Answer:

$\displaystyle \text{(i) } \frac{d}{dx} (f(x)) = \lim \limits_{h \to 0 } \frac{f(x+h) - f(x)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{\frac{2}{x+h} - \frac{2}{x}}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{2x - 2x - 2h}{hx(x+h)}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{- 2h}{hx(x+h)}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{- 2}{x(x+h)}$

$\displaystyle = \frac{-2}{x^2}$

$\\$

$\displaystyle \text{(ii) } \frac{d}{dx} (f(x)) = \lim \limits_{h \to 0 } \frac{f(x+h) - f(x)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ \frac{1}{ \sqrt{x+h} } - \frac{1}{\sqrt{x}} }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{\sqrt{x} - \sqrt{x+h}}{h \sqrt{x} \sqrt{x+h}} \times \frac{\sqrt{x} + \sqrt{x+h}}{\sqrt{x} + \sqrt{x+h}}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{x-x-h}{(h \sqrt{x} \sqrt{x+h})(\sqrt{x} + \sqrt{x+h})}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{-h}{(h \sqrt{x} \sqrt{x+h})(\sqrt{x} + \sqrt{x+h})}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{-1}{(\sqrt{x} \sqrt{x+h})(\sqrt{x} + \sqrt{x+h})}$

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

$\displaystyle = \frac{-1}{2x \sqrt{x}} = \frac{-1}{2x^{\frac{3}{2}}} = - \frac{1}{2} x^{-\frac{3}{2}}$

$\\$

$\displaystyle \text{(iii) } \frac{d}{dx} (f(x)) = \lim \limits_{h \to 0 } \frac{f(x+h) - f(x)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{\frac{1}{(x+h)^3} - \frac{1}{x^3} }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{x^3 - (x+h)^3 }{h(x+h)^3x^3}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{x^3 - x^3 - 3x^2 h - 3xh^2 - h^3 }{h(x+h)^3x^3}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ - 3x^2 h - 3xh^2 - h^3 }{h(x+h)^3x^3}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ h( - 3x^2 - 3xh - h^2) }{h(x+h)^3x^3}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ ( - 3x^2 - 3xh - h^2) }{(x+h)^3x^3}$

$\displaystyle = \frac{-3x^2}{x^6} = \frac{-3}{x^4} = -3 x^{-4}$

$\\$

$\displaystyle \text{(vi) } \frac{d}{dx} (f(x)) = \lim \limits_{h \to 0 } \frac{f(x+h) - f(x)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{\frac{(x+h)^2+1}{x+h} - \frac{x^2+1}{x} }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{\frac{x^2+2xh+h^2+1}{x+h} - \frac{x^2+1}{x} }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{x^3+2x^2h+h^2x+x-x^3-x^2h-x-h }{xh(x+h)}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{x^2h+h^2x-h }{x(x+h)}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ h(x^2+hx-1) }{x(x+h)}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ ( x^2+hx-1) }{x(x+h)}$

$\displaystyle = \frac{x^2-1}{x^2}$

$\\$

$\displaystyle \text{(v) } \frac{d}{dx} (f(x)) = \lim \limits_{h \to 0 } \frac{f(x+h) - f(x)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{\frac{(x+h)^2-1}{x+h} - \frac{x^2-1}{x} }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{\frac{x^2+2xh+h^2-1}{x+h} - \frac{x^2-1}{x} }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{x^3+2x^2h+h^2x-x-x^3-x^2h+x+h }{xh(x+h)}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{x^2h+h^2x+h }{x(x+h)}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ h(x^2+hx+1) }{x(x+h)}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ ( x^2+hx+1) }{x(x+h)}$

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

$\\$

$\displaystyle \text{(vi) } \frac{d}{dx} (f(x)) = \lim \limits_{h \to 0 } \frac{f(x+h) - f(x)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{\frac{x+h+1}{x+h+2} - \frac{x+1}{x+2} }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{(x+h+1)(x+2)-(x+h+2)(x+1)}{h(x+h+2)(x+2)}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{x^2+2x+hx+2h+x+2-x^2-x-hx-h-2x-2}{h(x+h+2)(x+2)}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{h }{h(x+h+2)(x+2)}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{1 }{(x+h+2)(x+2)}$

$\displaystyle = \frac{1}{(x+0+2)(x+2)} = \frac{1}{(x+2)^2}$

$\\$

$\displaystyle \text{(vii) } \frac{d}{dx} (f(x)) = \lim \limits_{h \to 0 } \frac{f(x+h) - f(x)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{\frac{x+h+2}{3(x+h)+5} - \frac{x+2}{3x+5} }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{\frac{x+h+2}{3x+3h+5} - \frac{x+2}{3x+5} }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{(x+h+2)(3x+5)-(3x+3h+5)(x+2)}{h(3x+3h+5)(3x+5)}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{3x^3+3xh+6x+5x+5h+10-3x^2-3xh-5x-6x-6h-10}{h(3x+3h+5)(3x+5)}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{-h }{h(3x+3h+5)(3x+5)}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{-1 }{(3x+3h+5)(3x+5)}$

$\displaystyle = \frac{-1}{(3x+5)^2}$

$\\$

$\displaystyle \text{(viii) } \frac{d}{dx} (f(x)) = \lim \limits_{h \to 0 } \frac{f(x+h) - f(x)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{k(x+h)^n-kx^n}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{k[(x+h)^n-x^n]}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{k[(x+h)^n-x^n]}{(x+h) - x}$

$\displaystyle \text{We know, } \lim \limits_{h \to 0 } \frac{x^m - a^m}{x-a} = ma^{m-1}$

$\displaystyle = knx^{n-1}$

$\\$

$\displaystyle \text{(ix) } \frac{d}{dx} (f(x)) = \lim \limits_{h \to 0 } \frac{f(x+h) - f(x)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{\frac{1}{\sqrt{3-x-h}} - \frac{1}{\sqrt{3-x}} }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ (\sqrt{3-x} - \sqrt{3-x-h}) }{h \sqrt{3-x} \sqrt{3-x-h} }$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ (\sqrt{3-x} - \sqrt{3-x-h}) }{h \sqrt{3-x} \sqrt{3-x-h} } \times \frac{\sqrt{3-x} + \sqrt{3-x-h}}{\sqrt{3-x} + \sqrt{3-x-h}}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{3-x-3+x+h}{h \sqrt{3-x} \sqrt{3-x-h} (\sqrt{3-x} + \sqrt{3-x-h})}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{h}{h \sqrt{3-x} \sqrt{3-x-h} (\sqrt{3-x} + \sqrt{3-x-h})}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{1}{ \sqrt{3-x} \sqrt{3-x-h} (\sqrt{3-x} + \sqrt{3-x-h})}$

$\displaystyle = \frac{1}{ \sqrt{3-x} \sqrt{3-x-0} (\sqrt{3-x} + \sqrt{3-x-0})}$

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

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

$\\$

$\displaystyle \text{(x) } \frac{d}{dx} (f(x)) = \lim \limits_{h \to 0 } \frac{f(x+h) - f(x)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{(x+h)^2+x+h+3-(x^2+x+3) }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{x^2+h^2+2xh+x+h+3-x^2-x-3 }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{h^2+2xh+h }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{h(h+2x+1) }{h}$

$\displaystyle = \lim \limits_{h \to 0 } h+2x+1$

$\displaystyle = 0+2x+1$

$\displaystyle = 2x+1$

$\\$

$\displaystyle \text{(xi) } \frac{d}{dx} (f(x)) = \lim \limits_{h \to 0 } \frac{f(x+h) - f(x)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{(x+h+2)^3-(x+2)^3 }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{x^3+3x^2h+3xh^2+h^3+6x^2+12xh+6h^2+12x+12h+8 -x^2-3x^2h-3xh^2-h^3 }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{3x^2h+3xh^2+h^3+12xh+6h^2+12h }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{h(3x^2+3xh+h^2+12x+6h+12) }{h}$

$\displaystyle = \lim \limits_{h \to 0 } 3x^2+3xh+h^2+12x+6h+12$

$\displaystyle = 3x^2+12x+12$

$\displaystyle = 3(x+2)^2$

$\\$

$\displaystyle \text{(xii) } \frac{d}{dx} (f(x)) = \lim \limits_{h \to 0 } \frac{f(x+h) - f(x)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{(x+h)^3+4(x+h)^2+3(x+h)+2 - x^3+4x^2+3x+2}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{3hx^2+3h^2x+h^3+8hx+4h^2+3h}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{h(3x^2+3hx+h^2+8x+4h+3) }{h}$

$\displaystyle = \lim \limits_{h \to 0 } 3x^2+3hx+h^2+8x+4h+3$

$\displaystyle = 3x^2+3(0)x+(0)^2+8x+4(0)+3$

$\displaystyle = 3x^2+8x+3$

$\\$

$\displaystyle \text{(xiii) } \frac{d}{dx} (f(x)) = \lim \limits_{h \to 0 } \frac{f(x+h) - f(x)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ ((x+h)^2+1)(x+h-5) - (x^2+1)(x-5) }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ 3hx^2+3h^2x-10hx+h^3-5h^2+h }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ h(3x^2+3hx-10x+h^2-5h+1 ) }{h}$

$\displaystyle = \lim \limits_{h \to 0 } 3x^2+3hx-10x+h^2-5h+1$

$\displaystyle = 3x^2-10x+1$

$\\$

$\displaystyle \text{(xiv) } \frac{d}{dx} (f(x)) = \lim \limits_{h \to 0 } \frac{f(x+h) - f(x)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{\sqrt{2(x+h)^2+1} - \sqrt{2x^2+1} }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{\sqrt{2x^2+h^2+4xh+1} - \sqrt{2x^2+1} }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{\sqrt{2x^2+h^2+4xh+1} - \sqrt{2x^2+1} }{h} \times \frac{\sqrt{2x^2+h^2+4xh+1} + \sqrt{2x^2+1}}{\sqrt{2x^2+h^2+4xh+1} + \sqrt{2x^2+1}}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{2x^2+2h^2+4xh+1 - 2x^2-1 }{h(\sqrt{2x^2+h^2+4xh+1} + \sqrt{2x^2+1})}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{h(2h+4x) }{h(\sqrt{2x^2+h^2+4xh+1} + \sqrt{2x^2+1})}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{(2h+4x) }{(\sqrt{2x^2+h^2+4xh+1} + \sqrt{2x^2+1})}$

$\displaystyle = \frac{(2(0)+4x) }{(\sqrt{2x^2+(0)^2+4x(0)+1} + \sqrt{2x^2+1})}$

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

$\\$

$\displaystyle \text{(xv) } \frac{d}{dx} (f(x)) = \lim \limits_{h \to 0 } \frac{f(x+h) - f(x)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ \frac{2(x+h)+3}{x+h-2} - \frac{2x+3}{x-2} }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ (2x+2h+3)(x-2)-(x+h-2)(2x+3) }{h(x+h-2)(x-2)}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ 2x^2+2xh+3x-4x-4h-6-2x^2-2xh+4x-3x-3h+6 }{h(x+h-2)(x-2)}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ -7h }{h(x+h-2)(x-2)}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ -7 }{(x+h-2)(x-2)}$

$\displaystyle = \frac{-7}{(x+0-2)(x-2) }$

$\displaystyle = \frac{-7}{(x-2)^2 }$

$\\$

Question 2: Differentiate each of the following from first principle:

$\displaystyle \text{(i) } e^{-x}$ $\displaystyle \text{(ii) } e^{3x}$ $\displaystyle \text{(iii) } e^{ax+b}$ $\displaystyle \text{(vi) } xe^{x}$

$\displaystyle \text{(v) } - x$ $\displaystyle \text{(vi) } (-x)^{-1}$ $\displaystyle \text{(vii) } \sin (x+1)$ $\displaystyle \text{(viii) } \cos \Big( x - \frac{\pi}{8} \Big)$

$\displaystyle \text{(ix) } x \sin x$ $\displaystyle \text{(x) } x \cos x$ $\displaystyle \text{(xi) } \sin ( 2x - 3)$

Answer:

$\displaystyle \text{(i) } \frac{d}{dx} (f(x)) = \lim \limits_{h \to 0 } \frac{f(x+h) - f(x)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{e^{-(x+h)} - e^{-x}}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{e^{-x} e^{-h} - e^{-x}}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{e^{-x} (e^{-h} - 1)}{h}$

$\displaystyle = - e^{-x} \lim \limits_{h \to 0 } \frac{(e^{-h} - 1)}{h}$

$\displaystyle = -e^{-x} (1) = -e^{-x}$

$\\$

$\displaystyle \text{(ii) } \frac{d}{dx} (f(x)) = \lim \limits_{h \to 0 } \frac{f(x+h) - f(x)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{e^{3(x+h)} - e^{3x}}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{e^{3x} e^{3h} - e^{3x}}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{e^{3x} (e^{3h} - 1)}{h}$

$\displaystyle = 3 e^{3x} \lim \limits_{h \to 0 } \frac{(e^{3h} - 1)}{3h}$

$\displaystyle = 3e^{3x} (1) = 3e^{3x}$

$\\$

$\displaystyle \text{(iii) } \frac{d}{dx} (f(x)) = \lim \limits_{h \to 0 } \frac{f(x+h) - f(x)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{e^{a(x+h)+b} - e^{ax+b}}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{e^{ax+b} e^{ah} - e^{ax+b}}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{e^{ax+b} (e^{ah} - 1)}{h}$

$\displaystyle = a e^{ax+b} \lim \limits_{h \to 0 } \frac{(e^{ah} - 1)}{ah}$

$\displaystyle = ae^{ax+b} (1) = ae^{ax+b}$

$\\$

$\displaystyle \text{(vi) } \frac{d}{dx} (f(x)) = \lim \limits_{h \to 0 } \frac{f(x+h) - f(x)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{(x+h)e^{(x+h)} - xe^{x}}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{(x+h)e^{x} e^{h} - xe^{x}}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{x e^x e^h + h e^x e^h - x e^x}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{x e^x e^h - x e^x}{h} + \lim \limits_{h \to 0 } \frac{h e^x e^h}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{x e^x (e^h - 1)}{h} + \lim \limits_{h \to 0 } e^x e^h$

$\displaystyle = x e^x + e^x$

$\\$

$\displaystyle \text{(v) } \frac{d}{dx} (f(x)) = \lim \limits_{h \to 0 } \frac{f(x+h) - f(x)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{-(x+h) - (-x)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{-x-h+x}{h}$

$\displaystyle = \lim \limits_{h \to 0 } -1$

$\displaystyle = \lim \limits_{h \to 0 } \frac{-h}{h}$

$\displaystyle = -1$

$\\$

$\displaystyle \text{(vi) } \frac{d}{dx} (f(x)) = \lim \limits_{h \to 0 } \frac{f(x+h) - f(x)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{\frac{1}{-(x+h)} - \frac{1}{-x} }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{-x+x+h}{hx(x+h)}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{1}{x(x+h)}$

$\displaystyle = \frac{1}{x \cdot x}$

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

$\\$

$\displaystyle \text{(vii) } \frac{d}{dx} (f(x)) = \lim \limits_{h \to 0 } \frac{f(x+h) - f(x)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{\sin ( x+h+1) - \sin (x+1)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{2 \cos (\frac{x+h+1+x+1}{2}) \sin (\frac{x+h+1-x-1}{2} )}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{2 \cos (\frac{2x+h+2}{2}) \sin (\frac{h}{2} )}{h}$

$\displaystyle = 2 \lim \limits_{h \to 0 } \frac{\cos (\frac{2x+h+2}{2}) \sin (\frac{h}{2} )}{\frac{h}{2}} \times \frac{1}{2}$

$\displaystyle = 2 \cos( x+1) \times \frac{1}{2}$

$\displaystyle = \cos ( x+1)$

$\\$

$\displaystyle \text{(viii) } \frac{d}{dx} (f(x)) = \lim \limits_{h \to 0 } \frac{f(x+h) - f(x)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{\cos ( x+h-\frac{\pi}{8}) - \cos (x-\frac{\pi}{8})}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{-2 \sin (\frac{x+h-\frac{\pi}{8}+x-\frac{\pi}{8}}{2}) \sin (\frac{x+h-\frac{\pi}{8}-x+\frac{\pi}{8}}{2} )}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{-2 \sin (\frac{2x+h-\frac{\pi}{4}}{2}) \sin (\frac{h}{2} )}{h}$

$\displaystyle = -2 \lim \limits_{h \to 0 } \frac{\sin (\frac{2x+h-\frac{\pi}{4}}{2}) \sin (\frac{h}{2} )}{\frac{h}{2}} \times \frac{1}{2}$

$\displaystyle = -2 \sin ( x-\frac{\pi}{8}) \times \frac{1}{2}$

$\displaystyle = -\sin ( x-\frac{\pi}{8})$

$\\$

$\displaystyle \text{(ix) } \frac{d}{dx} (f(x)) = \lim \limits_{h \to 0 } \frac{f(x+h) - f(x)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{(x+h) \sin (x+h) - x \sin x}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{(x+h) (\sin x \cos h + \cos x \sin h) - x \sin x}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{x\sin x \cos h + x\cos x \sin h + h\sin x \cos h + h\cos x \sin h- x \sin x}{h}$

$\displaystyle = x \sin x \lim \limits_{h \to 0 } \frac{\cos h - 1}{h} + x \cos x \lim \limits_{h \to 0 } \frac{\sin h}{h} + \sin x \lim \limits_{h \to 0 } \cos h + \cos x \lim \limits_{h \to 0 } \sin h$

$\displaystyle = x \sin x \lim \limits_{h \to 0 } \frac{-2 \sin^2 \frac{h}{2}}{\frac{h^2}{4}} \times \frac{h}{4} + x \cos x (1) + \sin x (1) + \cos x (0)$

$\displaystyle = -2x \sin x \times \lim \limits_{h \to 0 } \frac{h}{4} + x \cos x + \sin x$

$\displaystyle = x \cos x + \sin x$

$\\$

$\displaystyle \text{(x) } \frac{d}{dx} (f(x)) = \lim \limits_{h \to 0 } \frac{f(x+h) - f(x)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{(x+h) \cos (x+h) - x \cos x}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{(x+h) (\cos x \cos h - \sin x \sin h) - x \cos x}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{x\cos x \cos h - x\sin x \sin h + h\cos x \cos h - h\sin x \sin h- x \cos x}{h}$

$\displaystyle = x \cos x \lim \limits_{h \to 0 } \frac{\cos h - 1}{h} - x \sin x \lim \limits_{h \to 0 } \frac{\sin h}{h} + \cos x \lim \limits_{h \to 0 } \cos h + \sin x \lim \limits_{h \to 0 } \sin h$

$\displaystyle = x \cos x \lim \limits_{h \to 0 } \frac{-2 \sin^2 \frac{h}{2}}{\frac{h^2}{4}} \times \frac{h}{4} - x \sin x (1) + \cos x (1) + \sin x (0)$

$\displaystyle = -x \cos x \times (0) - x \sin x + \cos x$

$\displaystyle = - x \sin x + \cos x$

$\\$

$\displaystyle \text{(xi) } \frac{d}{dx} (f(x)) = \lim \limits_{h \to 0 } \frac{f(x+h) - f(x)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{\sin(2x+2h-3) - \sin(2x-3)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ 2 \cos (\frac{2x+2h-3+2x-3}{2} ) \sin( \frac{2x+2h-3+2x-3}{2} ) }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ 2 \cos (\frac{4x+2h-6}{2} ) \sin( h) }{h}$

$\displaystyle = \lim \limits_{h \to 0 } 2 \cos \Big( \frac{4x+2h-6}{2} \Big) \lim \limits_{h \to 0 } \frac{\sin h}{h}$

$\displaystyle = 2 \cos \Big( \frac{4x-6}{2} \Big) (1)$

$\displaystyle = 2 \cos (2x-3)$

$\\$

Question 3: Differentiate each of the following from first principle:

$\displaystyle \text{(i) } \sqrt{\sin 2x }$ $\displaystyle \text{(ii) } \frac{\sin x}{x}$ $\displaystyle \text{(iii) } \frac{\cos x}{x}$

$\displaystyle \text{(vi) } x^2 \sin x$ $\displaystyle \text{(v) } \sqrt{\sin (3x+1) }$ $\displaystyle \text{(vi) } \sin x + \cos x$

$\displaystyle \text{(vii) } x^2 e^x$ $\displaystyle \text{(viii) } e^{x^2 + 1 }$ $\displaystyle \text{(ix) } e^{\sqrt{x} }$

$\displaystyle \text{(x) } e^{ \sqrt{ax+b}}$ $\displaystyle \text{(xi) } a^{x}$ $\displaystyle \text{(xii) } 3^{x^2}$

Answer:

$\displaystyle \text{(i) } \frac{d}{dx} (f(x)) = \lim \limits_{h \to 0 } \frac{f(x+h) - f(x)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{\sqrt{\sin(2x+2h)} - \sqrt{\sin 2x} }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{\sqrt{\sin(2x+2h)} - \sqrt{\sin 2x} }{h} \times \frac{\sqrt{\sin(2x+2h)} + \sqrt{\sin 2x}}{\sqrt{\sin(2x+2h)} + \sqrt{\sin 2x}}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{\sin(2x+2h) - \sin 2x }{h(\sqrt{\sin(2x+2h)} + \sqrt{\sin 2x})}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ 2 \cos (\frac{2x+2h+2x}{2} ) \sin (\frac{2x+2h-2x}{2} ) }{h(\sqrt{\sin(2x+2h)} + \sqrt{\sin 2x})}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ 2 \cos ( 2x+h) \sin h }{h(\sqrt{\sin(2x+2h)} + \sqrt{\sin 2x})}$

$\displaystyle = \lim \limits_{h \to 0 } 2 \cos( 2x+h) \ \lim \limits_{h \to 0 } \frac{\sin h}{h} \ \lim \limits_{h \to 0 } \frac{1}{(\sqrt{\sin(2x+2h)} + \sqrt{\sin 2x})}$

$\displaystyle = 2 \cos 2x ( 1) \times \frac{1}{\sqrt{\sin 2x} + \sqrt{\sin 2x}}$

$\displaystyle = 2 \cos 2x ( 1) \times \frac{1}{2\sqrt{\sin 2x} }$

$\displaystyle = \frac{\cos 2x}{\sqrt{\sin 2x}}$

$\\$

$\displaystyle \text{(ii) } \frac{d}{dx} (f(x)) = \lim \limits_{h \to 0 } \frac{f(x+h) - f(x)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{\frac{\sin ( x+h)}{x+h} - \frac{\sin x }{x} }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{x \sin ( x+h) - ( x+h) \sin x }{hx(x+h)}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{x (\sin x \cos h + \cos x \sin h ) - x \sin x - h \sin x }{hx(x+h)}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{x\sin x \cos h + x\cos x \sin h - x \sin x - h \sin x }{hx(x+h)}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{x\sin x \cos h - x \sin x + x\cos x \sin h - h \sin x }{hx(x+h)}$

$\displaystyle = x \sin x \lim \limits_{h \to 0 } \frac{\cos h - 1}{h} + \frac{x \cos x}{x} \lim \limits_{h \to 0 } \frac{\sin h}{h} \lim \limits_{h \to 0 }\frac{1}{x+h} - \frac{\sin x }{x} \lim \limits_{h \to 0 } \frac{1}{x+h}$

$\displaystyle = x \sin x \lim \limits_{h \to 0 } \frac{-2 \sin^2 \frac{h}{2}}{h} + \frac{x \cos x}{x} \lim \limits_{h \to 0 } \frac{\sin h}{h} \lim \limits_{h \to 0 }\frac{1}{x+h} - \frac{\sin x }{x} \lim \limits_{h \to 0 } \frac{1}{x+h}$

$\displaystyle = x \sin x \lim \limits_{h \to 0 } \frac{-2 \sin^2 \frac{h}{2}}{\frac{h^2}{4}} \times \frac{h}{4} + \frac{x \cos x}{x} \lim \limits_{h \to 0 } \frac{\sin h}{h} \lim \limits_{h \to 0 }\frac{1}{x+h} - \frac{\sin x }{x} \lim \limits_{h \to 0 } \frac{1}{x+h}$

$\displaystyle = -x \sin x \lim \limits_{h \to 0 } \frac{h}{2} + \frac{x \cos x}{x} \lim \limits_{h \to 0 } \frac{\sin h}{h} \lim \limits_{h \to 0 }\frac{1}{x+h} - \frac{\sin x }{x} \lim \limits_{h \to 0 } \frac{1}{x+h}$

$\displaystyle = - x \sin x (\frac{1}{2}) (0) + \frac{\cos x } {x} - \frac{\sin x}{x^2}$

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

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

$\\$

$\displaystyle \text{(iii) } \frac{d}{dx} (f(x)) = \lim \limits_{h \to 0 } \frac{f(x+h) - f(x)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{\frac{\cos ( x+h)}{x+h} - \frac{\cos x }{x} }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{x \cos ( x+h) - ( x+h) \cos x }{hx(x+h)}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{x (\cos x \cos h - \sin x \sin h ) - x \cos x - h \cos x }{hx(x+h)}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{x\cos x \cos h - x\sin x \sin h - x \cos x - h \cos x }{hx(x+h)}$

$\displaystyle = \frac{x\cos x \cos h - x \cos x - x\sin x \sin h - h \cos x }{hx(x+h)}$

$\displaystyle = x \cos x \lim \limits_{h \to 0 } \frac{\cos h - 1}{h} - \frac{x \sin x}{x} \lim \limits_{h \to 0 } \frac{\sin h}{h} \lim \limits_{h \to 0 } \frac{1}{x+h} - \frac{\cos x}{x} \lim \limits_{h \to 0 } \frac{1}{x+h}$

$\displaystyle = x \cos x \lim \limits_{h \to 0 } \frac{-2 \sin^2 \frac{h}{2}}{\frac{h^2}{4}} \times \frac{h}{4} - \frac{x \sin x}{x} \lim \limits_{h \to 0 } \frac{\sin h}{h} \lim \limits_{h \to 0 } \frac{1}{x+h} - \frac{\cos x}{x} \lim \limits_{h \to 0 } \frac{1}{x+h}$

$\displaystyle = -x \cos x \lim \limits_{h \to 0 } \frac{h}{4} - \frac{x \sin x}{x} \lim \limits_{h \to 0 } \frac{\sin h}{h} \lim \limits_{h \to 0 } \frac{1}{x+h} - \frac{\cos x}{x} \lim \limits_{h \to 0 } \frac{1}{x+h}$

$\displaystyle = 0 - \frac{\sin x}{x} - \frac{\cos x}{x^2}$

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

$\\$

$\displaystyle \text{(vi) } \frac{d}{dx} (f(x)) = \lim \limits_{h \to 0 } \frac{f(x+h) - f(x)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{(x+h)^2 \sin ( x+ h) - x^2 \sin x}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{(x^2+h^2+ 2xh) (\sin x \cos h+ \cos x \sin h) - x^2 \sin x}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ x^2 \sin x \cos h+ x^2 \cos x \sin h + h^2\sin x \cos h+ h^2 \cos x \sin h + 2xh\sin x \cos h+ 2xh\cos x \sin h- x^2 \sin x}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ x^2 \sin x \cos h - x^2 \sin x + x^2 \cos x \sin h + h^2\sin x \cos h+ h^2 \cos x \sin h + 2xh\sin x \cos h+ 2xh\cos x \sin h}{h}$

$\displaystyle = x^2 \sin x \lim \limits_{h \to 0 } \frac{\cos h - 1}{h} + x^2 \cos x \lim \limits_{h \to 0 } \frac{\sin h}{h} + \sin x \lim \limits_{h \to 0 } h \cos h \\ \\ + \cos x \lim \limits_{h \to 0 }h \sin h + 2x \sin x \lim \limits_{h \to 0 } \cos h + 2x \cos x \lim \limits_{h \to 0 } \sin h$

$\displaystyle = x^2 \sin x \lim \limits_{h \to 0 } \frac{-2 \sin^2 \frac{h}{2}}{\frac{h^2}{4}} \times \frac{h}{4}+ x^2 \cos x \lim \limits_{h \to 0 } \frac{\sin h}{h} + \sin x \lim \limits_{h \to 0 } h \cos h \\ \\ + \cos x \lim \limits_{h \to 0 }h \sin h + 2x \sin x \lim \limits_{h \to 0 } \cos h + 2x \cos x \lim \limits_{h \to 0 } \sin h$

$\displaystyle = -x^2 \sin x \lim \limits_{h \to 0 } \frac{h}{2}+ x^2 \cos x \lim \limits_{h \to 0 } \frac{\sin h}{h} + \sin x \lim \limits_{h \to 0 } h \cos h \\ \\ + \cos x \lim \limits_{h \to 0 }h \sin h + 2x \sin x \lim \limits_{h \to 0 } \cos h + 2x \cos x \lim \limits_{h \to 0 } \sin h$

$\displaystyle = - x^2 \sin x (0) + x^2 \cos x (1) + \sin x (0) + \cos x ( 0) + 2x \sin x (1) + 2x \cos (x)$

$\displaystyle = 0 + x^2 \cos x + 2x \sin x$

$\displaystyle = x^2 \cos x + 2x \sin x$

$\\$

$\displaystyle \text{(v) } \frac{d}{dx} (f(x)) = \lim \limits_{h \to 0 } \frac{f(x+h) - f(x)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ \sqrt{\sin ( 3( x+h)+1)} - \sqrt{\sin ( 3x+1)} }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ \sqrt{\sin ( 3( x+h)+1)} - \sqrt{\sin ( 3x+1)} }{h} \times \frac{ \sqrt{\sin ( 3( x+h)+1)} + \sqrt{\sin ( 3x+1)}}{ \sqrt{\sin ( 3( x+h)+1)} + \sqrt{\sin ( 3x+1)}}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{\sin(3x+3h+1) - \sin ( 3x+1)}{h(\sqrt{\sin ( 3x+3h+1)} + \sqrt{\sin ( 3x+1)})}$

$\\$

$\displaystyle \text{(vi) } \frac{d}{dx} (f(x)) = \lim \limits_{h \to 0 } \frac{f(x+h) - f(x)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ \sin (x+h) + \cos ( x+h) - \sin x - \cos x}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{\sin ( x+ h) - \sin x}{h} + \lim \limits_{h \to 0 } \frac{\cos ( x+ h) - \cos x}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{2 \cos (\frac{2x+h}{2}) \sin \frac{h}{2}}{h} + \lim \limits_{h \to 0 } \frac{ - 2 \sin (\frac{2x+h}{2}) \sin \frac{h}{2}}{h}$

$\displaystyle = 2 \lim \limits_{h \to 0 } \cos \Big( \frac{2x+h}{2} \Big) \lim \limits_{h \to 0 } \frac{\sin \frac{h}{2}}{\frac{h}{2}} \times \frac{1}{2} - 2 \lim \limits_{h \to 0 } \sin \Big( \frac{2x+h}{2} \Big) \lim \limits_{h \to 0 }\frac{\sin \frac{h}{2}}{2} \times \frac{1}{2}$

$\displaystyle = 2\cos x \times \frac{1}{2} - 2 \sin x \times \frac{1}{2}$

$\displaystyle = \cos x - \sin x$

$\\$

$\displaystyle \text{(vii) } \frac{d}{dx} (f(x)) = \lim \limits_{h \to 0 } \frac{f(x+h) - f(x)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ (x+h)^2 e^{x+h} - x^2 e^x }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ (x^2 + h^2 + 2xh) e^x e^h - x^2 e^x }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ x^2 e^x e^h + h^2 e^x e^h + 2xh e^x e^h - x^2 e^x }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{x^2 e^x e^h- x^2 e^x}{h} +\lim \limits_{h \to 0 } \frac{2xh e^x e^h }{h} +\lim \limits_{h \to 0 } \frac{h^2 e^x e^h}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{x^2 e^x (e^h- 1)}{h} +\lim \limits_{h \to 0 } 2x e^x e^h +\lim \limits_{h \to 0 } h e^x e^h$

$\displaystyle = x^2 e^x + 2 x e^x$

$\displaystyle = (x^2 + 2x) e^x$

$\\$

$\displaystyle \text{(viii) } \frac{d}{dx} (f(x)) = \lim \limits_{h \to 0 } \frac{f(x+h) - f(x)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ e^{(x+h)^2+1} - e^{x^2+1}}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ e^{x^2+h^2+2hx+1} - e^{x^2+1}}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ e^{x^2+1} e^{h^2+2hx} - e^{x^2+1}}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ e^{x^2+1} (e^{h^2+2hx} - 1)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ e^{x^2+1} (e^{h(h+2x)} - 1)}{h(h+2x)} \times (h+2x)$

$\displaystyle = e^{x^2+1} \lim \limits_{h \to 0 } \frac{ (e^{h(h+2x)} - 1)}{h(h+2x)} \times \lim \limits_{h \to 0 } (h+2x)$

$\displaystyle = e^{x^2+1} (1) (2x)$

$\displaystyle = 2xe^{x^2+1}$

$\\$

$\displaystyle \text{(ix) } \frac{d}{dx} (f(x)) = \lim \limits_{h \to 0 } \frac{f(x+h) - f(x)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ e^{\sqrt{ 2(x+h)}} - e^{\sqrt{2x}} }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ e^{\sqrt{ 2(x+h)}} - e^{\sqrt{2x}} }{2x+2h - 2x}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ e^{\sqrt{ 2(x+h)}} - e^{\sqrt{2x}} }{ (\sqrt{2x+2h})^2 - (\sqrt{2x})^2}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ e^{\sqrt{ 2(x+h)}} - e^{\sqrt{2x}} }{ (\sqrt{2x+2h} - \sqrt{2x})(\sqrt{2x+2h} + \sqrt{2x})}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ e^{\sqrt{2x}} (e^{\sqrt{ 2(x+h)} - \sqrt{2x} } - 1) }{ (\sqrt{2x+2h} - \sqrt{2x})(\sqrt{2x+2h} + \sqrt{2x})}$

$\displaystyle = e^{\sqrt{2x}} \lim \limits_{h \to 0 } \frac{ e^{\sqrt{ 2(x+h)} - \sqrt{2x} } - 1}{ \sqrt{2x+2h} - \sqrt{2x} } \times \lim \limits_{h \to 0 } \ \frac{1}{ ( \sqrt{2x+2h} + \sqrt{2x} ) }$

$\displaystyle = e^{\sqrt{2x}} \times \frac{1}{ ( \sqrt{2x} + \sqrt{2x} ) }$

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

$\\$

$\displaystyle \text{(x) } \frac{d}{dx} (f(x)) = \lim \limits_{h \to 0 } \frac{f(x+h) - f(x)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{e^{ \sqrt{a(x+h)+b}} - e^{ \sqrt{ax+b}} }{h}$

$\displaystyle = a\lim \limits_{h \to 0 } \frac{e^{ \sqrt{ax+b+ah}} - e^{ \sqrt{ax+b}} }{(ax+ah+b)-(ax+b)}$

$\displaystyle = a\lim \limits_{h \to 0 } \frac{e^{ \sqrt{ax+b}} ( e^{ \sqrt{ax+b+ah} - \sqrt{ax+b} } - 1) }{(ax+ah+b)-(ax+b)}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{e^{ \sqrt{ax+b}} ( e^{ \sqrt{ax+b+ah} - \sqrt{ax+b} } - 1) }{(\sqrt{ax+ah+b})^2 - (\sqrt{ax+b})^2}$

$\\$

$\displaystyle \text{(xi) } \frac{d}{dx} (f(x)) = \lim \limits_{h \to 0 } \frac{f(x+h) - f(x)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{a^{\sqrt{x+h}} - a^{\sqrt{x}}}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{a^{\sqrt{x}}(a^{\sqrt{x+h} - \sqrt{x}} - 1)}{(x+h)-(x)}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{a^{\sqrt{x}}(a^{\sqrt{x+h} - \sqrt{x}} - 1)}{(\sqrt{x+h})^2-(\sqrt{x})^2}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{a^{\sqrt{x}}(a^{\sqrt{x+h} - \sqrt{x}} - 1)}{(\sqrt{x+h})-(\sqrt{x})(\sqrt{x+h})+(\sqrt{x})}$

$\displaystyle = a^{\sqrt{x}} \lim \limits_{h \to 0 } \frac{(a^{\sqrt{x+h} - \sqrt{x}} - 1)}{(\sqrt{x+h})-(\sqrt{x})} \times \lim \limits_{h \to 0 } \frac{1}{(\sqrt{x+h})+(\sqrt{x})}$

$\displaystyle = a^{\sqrt{x}} \log_e a \frac{1}{2\sqrt{x}}$

$\displaystyle = \frac{1}{2\sqrt{x}} a^{\sqrt{x}} \log_e a$

$\\$

$\displaystyle \text{(xii) } \frac{d}{dx} (f(x)) = \lim \limits_{h \to 0 } \frac{f(x+h) - f(x)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ 3^{(x+h)^2} -3^{x^2} }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ 3^{x^2+h^2+2xh} -3^{x^2} }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ 3^{x^2}(3^{h^2+2xh} -1) }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ 3^{x^2}(3^{h(h+2x)} -1) }{h(h+2x)} \times ( h+2x)$

$\displaystyle = 3^{x^2} \log 3 ( 2x)$

$\displaystyle = 2x 3^{x^2} \log 3$

$\\$

Question 4: Differentiate each of the following from first principle:

$\displaystyle \text{(i) } \tan^2 x$ $\displaystyle \text{(ii) } \tan ( 2x+1)$ $\displaystyle \text{(iii) } \tan 2x$ $\displaystyle \text{(vi) } \sqrt{\tan x}$

Answer:

$\displaystyle \text{(i) } \frac{d}{dx} (f(x)) = \lim \limits_{h \to 0 } \frac{f(x+h) - f(x)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ \tan^2 ( x+h) - \tan^2 x }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ [\tan ( x+h) - \tan x][\tan ( x+h) + \tan x] }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ [ \frac{\sin ( x+h) }{\cos ( x+h)} + \frac{\sin x}{\cos x} ][ \frac{\sin ( x+h) }{ \cos ( x+h) } - \frac{\sin x}{\cos x} ] }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{[ \sin ( x+h) \cos x + \cos ( x+h) \sin x ] [\sin ( x+h) \cos x - \cos ( x+h) \sin x ]}{h \cos^2 x \cos^2 ( x+h) }$

$\displaystyle = \lim \limits_{h \to 0 } \frac{\sin ( 2x + h) \sin h }{h \cos^2 x \cos^2 ( x+h)}$

$\displaystyle = \frac{1}{\cos^2 x} \lim \limits_{h \to 0 } \ \sin ( 2x+h) \lim \limits_{h \to 0 } \ \frac{\sin h}{h} \lim \limits_{h \to 0 } \ \frac{1}{\cos^2 (x+h)}$

$\displaystyle = \frac{1}{\cos^2 x} \sin 2x (1) \frac{1}{\cos^2 x}$

$\displaystyle = \frac{1}{\cos^2 x} 2 \sin x \cos x \frac{1}{\cos^2 x}$

$\displaystyle = 2 \frac{\sin x }{\cos x} \times \frac{1}{\cos^2 x}$

$\displaystyle = 2 \tan x \sec^2 x$

$\\$

$\displaystyle \text{(ii) } \frac{d}{dx} (f(x)) = \lim \limits_{h \to 0 } \frac{f(x+h) - f(x)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ \tan ( 2x+2h+1) - \tan (2x+1) }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ \frac{\sin ( 2x+2h+1) }{\cos ( 2x+2h+1)} - \frac{\sin ( 2x+1)}{\cos (2x+1)} }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{\sin ( 2x+2h+1) \cos ( 2x+1) - \cos ( 2x+2h+1) \sin ( 2x+1)}{h \cos ( 2x+2h+1) \cos ( 2x+1) }$

$\displaystyle = \lim \limits_{h \to 0 } \frac{\sin ( 2x+2h+1-2x-1)}{h \cos ( 2x+2h+1) \cos ( 2x+1)}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{\sin ( 2h)}{h \cos ( 2x+2h+1) \cos ( 2x+1)}$

$\displaystyle = \frac{1}{\cos (2x+1)} \ \lim \limits_{h \to 0 } \ \frac{\sin 2h}{2h} \times 2 \lim \limits_{h \to 0 } \ \frac{1}{\cos ( 2x + 2h + 1)}$

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

$\displaystyle = 2 \sec^2 (2x+1)$

$\\$

$\displaystyle \text{(iii) } \frac{d}{dx} (f(x)) = \lim \limits_{h \to 0 } \frac{f(x+h) - f(x)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ \tan ( 2x+2h) - \tan (2x) }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ \frac{\sin ( 2x+2h) }{\cos ( 2x+2h)} - \frac{\sin ( 2x)}{\cos (2x)} }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{\sin ( 2x+2h) \cos ( 2x) - \cos ( 2x+2h) \sin ( 2x)}{h \cos ( 2x+2h) \cos ( 2x) }$

$\displaystyle = \lim \limits_{h \to 0 } \frac{\sin ( 2x+2h-2x)}{h \cos ( 2x+2h) \cos ( 2x)}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{\sin ( 2h)}{h \cos ( 2x+2h) \cos ( 2x)}$

$\displaystyle = \frac{1}{\cos (2x)} \ \lim \limits_{h \to 0 } \ \frac{\sin 2h}{2h} \times 2 \lim \limits_{h \to 0 } \ \frac{1}{\cos ( 2x + 2h)}$

$\displaystyle = \frac{2}{\cos^2 ( 2x)}$

$\displaystyle = 2 \sec^2 (2x)$

$\\$

$\displaystyle \text{(iv) } \frac{d}{dx} (f(x)) = \lim \limits_{h \to 0 } \frac{f(x+h) - f(x)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ \sqrt{\tan ( x+h)} - \sqrt{\tan x} }{h} \times \frac{\sqrt{\tan ( x+h)} + \sqrt{\tan x}}{\sqrt{\tan ( x+h)} + \sqrt{\tan x}}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{\tan ( x+h) - \tan x}{h( \sqrt{\tan ( x+h)} + \sqrt{\tan x})}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{\frac{\sin (x+h)}{\cos (x+h)} - \frac{\sin x}{\cos x} }{h( \sqrt{\tan ( x+h)} + \sqrt{\tan x})}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ \sin ( x+ h) \cos x - \cos (x+h) \sin x }{h( \sqrt{\tan ( x+h)} + \sqrt{\tan x}) \cos (x+h) \cos x}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{\sin h }{h( \sqrt{\tan ( x+h)} + \sqrt{\tan x}) \cos (x+h) \cos x}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{\sin h }{h} \lim \limits_{h \to 0 } \ \frac{1}{( \sqrt{\tan ( x+h)} + \sqrt{\tan x}) \cos (x+h) \cos x}$

$\displaystyle = (1) \frac{1}{2 \sqrt{\tan x} \cos^2 x}$

$\displaystyle = \frac{\sec^2 x}{2 \sqrt{\tan x} }$

$\\$

Question 5: Differentiate each of the following from first principle:

$\displaystyle \text{(i) } \sin \sqrt{2x}$ $\displaystyle \text{(ii) } \cos \sqrt{x}$ $\displaystyle \text{(iii) } \tan \sqrt{x}$ $\displaystyle \text{(vi) } \tan x^2$

Answer:

$\displaystyle \text{(i) } \frac{d}{dx} (f(x)) = \lim \limits_{h \to 0 } \frac{f(x+h) - f(x)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ \sin \sqrt{2x+2h} - \sin \sqrt{2x} }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ 2 \sin ( \frac{\sqrt{2x+2h}-\sqrt{2x} }{2} ) \cos ( \frac{\sqrt{2x+2h}+\sqrt{2x}}{2} ) }{h}$

$\displaystyle = 2 \lim \limits_{h \to 0 } \frac{ 2 \sin ( \frac{\sqrt{2x+2h}-\sqrt{2x} }{2} ) \cos ( \frac{\sqrt{2x+2h}+\sqrt{2x}}{2} ) }{2x+2h - 2x}$

$\displaystyle = 2 \lim \limits_{h \to 0 } \frac{ 2 \sin ( \frac{\sqrt{2x+2h}-\sqrt{2x} }{2} ) \cos ( \frac{\sqrt{2x+2h}+\sqrt{2x}}{2} ) }{( \sqrt{2x+2h})^2 - (\sqrt{2x})^2}$

$\displaystyle = 2 \lim \limits_{h \to 0 } \frac{ 2 \sin ( \frac{\sqrt{2x+2h}-\sqrt{2x} }{2} ) \cos ( \frac{\sqrt{2x+2h}+\sqrt{2x}}{2} ) }{( \sqrt{2x+2h} - \sqrt{2x})( \sqrt{2x+2h} + \sqrt{2x}) }$

$\displaystyle = \lim \limits_{h \to 0 } \frac{\sin ( \frac{\sqrt{2x+2h}-\sqrt{2x} }{2} )}{\frac{( \sqrt{2x+2h} - \sqrt{2x})}{2} } \times \lim \limits_{h \to 0 } \frac{\cos ( \frac{\sqrt{2x+2h}+\sqrt{2x}}{2} )}{\frac{( \sqrt{2x+2h} + \sqrt{2x})}{2}}$

$\displaystyle = 1 \times \frac{\cos \sqrt{2x} }{ \sqrt{2x}}$

$\displaystyle = \frac{\cos \sqrt{2x} }{ \sqrt{2x}}$

$\\$

$\displaystyle \text{(ii) } \frac{d}{dx} (f(x)) = \lim \limits_{h \to 0 } \frac{f(x+h) - f(x)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ \cos \sqrt{x+h} - \cos \sqrt{x} }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ -2 \sin ( \frac{\sqrt{x+h}+\sqrt{x} }{2} ) \sin ( \frac{\sqrt{x+h}-\sqrt{x}}{2} ) }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ -2 \sin ( \frac{\sqrt{x+h}+\sqrt{x} }{2} ) \sin ( \frac{\sqrt{x+h}-\sqrt{x}}{2} ) }{x+h - x}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ -2 \sin ( \frac{\sqrt{x+h}+\sqrt{x} }{2} ) \sin ( \frac{\sqrt{x+h}-\sqrt{x}}{2} ) }{(\sqrt{x+h})^2 - (\sqrt{x})^2}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ -2 \sin ( \frac{\sqrt{x+h}+\sqrt{x} }{2} ) \sin ( \frac{\sqrt{x+h}-\sqrt{x}}{2} ) }{(\sqrt{x+h} - \sqrt{x})(\sqrt{x+h} + \sqrt{x})}$

$\displaystyle = 1 \lim \limits_{h \to 0 } \frac{\sin ( \frac{\sqrt{x+h}-\sqrt{x}}{2} ) }{ \frac{\sqrt{x+h}-\sqrt{x}}{2} } \times \lim \limits_{h \to 0 } \frac{- \sin ( \frac{\sqrt{x+h}+\sqrt{x} }{2} ) }{\sqrt{x+h}+\sqrt{x}}$

$\displaystyle = \frac{- \sin \sqrt{x}}{2 \sqrt{x}}$

$\\$

$\displaystyle \text{(iii) } \frac{d}{dx} (f(x)) = \lim \limits_{h \to 0 } \frac{f(x+h) - f(x)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ \tan \sqrt{x+h} - \tan \sqrt{x} }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ \sin ( \sqrt{x+h} - \sqrt{x} ) }{h \cos \sqrt{x+h} \cos \sqrt{x} }$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ \sin ( \sqrt{x+h} - \sqrt{x} ) }{(x+h-x) \cos \sqrt{x+h} \cos \sqrt{x} }$

$\displaystyle = \lim \limits_{h \to 0 } \frac{ \sin ( \sqrt{x+h} - \sqrt{x} ) }{(\sqrt{x+h} - \sqrt{x})(\sqrt{x+h} + \sqrt{x}) \cos \sqrt{x+h} \cos \sqrt{x} }$

$\displaystyle = \lim \limits_{h \to 0 } \frac{\sin ( \sqrt{x+h} - \sqrt{x} )}{(\sqrt{x+h} - \sqrt{x})} \times \lim \limits_{h \to 0 } \frac{1}{(\sqrt{x+h} + \sqrt{x}) \cos \sqrt{x+h} \cos \sqrt{x}}$