# Differentiation based on first Principle

Example : Using the first principle find the differential coefficient of y = f (x), where f (x) is

(i) ex

(ii) tan x

(iii)$\displaystyle \frac{sinx}{x}$

(iv) tan−1 x

Solution :

(i) y = f (x) = ex

we have $\displaystyle f'(x) = \lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}$

$\displaystyle = \lim_{h\rightarrow 0} \frac{e^{x+h}-e^x}{h}$

$\displaystyle = \lim_{h\rightarrow 0} \frac{e^x(e^h -1)}{h}$

$\displaystyle = e^x \lim_{h\rightarrow 0} \frac{(e^h -1)}{h}$

$\displaystyle = e^x \times 1$

= ex

(ii) y = f (x) = tan x

We have , $\displaystyle f'(x) = \lim_{h\rightarrow 0} \frac{f(x+h) - f(x)}{h}$

$\displaystyle f'(x) = \lim_{h\rightarrow 0} \frac{tan(x+h) - tanx }{h}$

$\displaystyle = \lim_{h\rightarrow 0} \frac{\frac{sin(x+h)}{cos(x+h)} - \frac{sinx}{cosx}}{h}$

$\displaystyle = \lim_{h\rightarrow 0} \frac{sin(x+h)cosx-cos(x+h)sinx}{h cos(x+h)cosx}$

$\displaystyle = \lim_{h\rightarrow 0} \frac{sin(x+h-x)}{h cos(x+h)cosx}$

$\displaystyle = \lim_{h\rightarrow 0} \frac{sinh}{h cos(x+h)cosx}$

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

= sec2x

(iii) y = f (x) = sinx/x

We have , $\displaystyle f'(x) = \lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}$

$\displaystyle = \lim_{h\rightarrow 0} \frac{\frac{sin(x+h)}{x+h} - \frac{sinx}{x}}{h}$

$\displaystyle = \lim_{h\rightarrow 0} \frac{x sin(x+h) - (x+h)sinx}{h (x+h)x}$

$\displaystyle = \lim_{h\rightarrow 0} \frac{x (sin(x+h) -sinx )-h sinx}{h (x+h)x}$

$\displaystyle = \lim_{h\rightarrow 0} \frac{x (2 cos(x+h/2) .sinh/2 )-h sinx}{h (x+h)x}$

$\displaystyle = \lim_{h\rightarrow 0} \frac{x (cos(x+h/2 ) .sinh/2 )}{(h/2) (x+h)x}-\lim_{h\rightarrow 0} \frac{sinx}{(x+h)x}$

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

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

(iv) y = f (x) = tan−1 x

we have , $\displaystyle f'(x) = \lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}$

$\displaystyle = \lim_{h\rightarrow 0} \frac{tan^{-1}(x+h)- tan^{-1}x}{h}$

$\displaystyle = \lim_{h\rightarrow 0} \frac{tan^{-1}\frac{x+h-x}{1+x(x+h)}}{h}$

$\displaystyle = \lim_{h\rightarrow 0} \frac{tan^{-1}\frac{h}{1+x(x+h)}}{\frac{h(1+x(x+h))}{1+x(x+h)}}$

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

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