Find the derivation of each of the following from the first principle:

Question:

Find the derivation of each of the following from the first principle:

$\tan ^{2} x$

 

Solution:

Let $f(x)=\tan ^{2} x$

We need to find the derivative of f(x) i.e. f’(x)

We know that,

$\mathrm{f}^{\prime}(\mathrm{x})=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(\mathrm{x}+\mathrm{h})-\mathrm{f}(\mathrm{x})}{\mathrm{h}}$ …(i)

$f(x)=\tan ^{2} x$

$f(x+h)=\tan ^{2}(x+h)$

Putting values in (i), we get

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{\tan ^{2}(x+h)-\tan ^{2} x}{h}$

$=\lim _{h \rightarrow 0} \frac{[\tan (x+h)-\tan x][\tan (x+h)+\tan x]}{h}$

Using:

$\tan x=\frac{\sin x}{\cos x}$

$=\lim _{h \rightarrow 0} \frac{\left[\frac{\sin (x+h)}{\cos (x+h)}-\frac{\sin x}{\cos x}\right]\left[\frac{\sin (x+h)}{\cos (x+h)}+\frac{\sin x}{\cos x}\right]}{h}$

$=\lim _{h \rightarrow 0} \frac{\left[\frac{\sin (x+h) \cos x-\sin x \cos (x+h)}{\cos (x+h) \cos x}\right]\left[\frac{\sin (x+h) \cos x+\sin x \cos (x+h)}{\cos (x+h) \cos x}\right]}{h}$

$=\lim _{h \rightarrow 0} \frac{\{\sin [(x+h)-x]\}\{\sin [(x+h)+x]\}}{h\left[\cos ^{2}(x+h) \cos ^{2} x\right]}$

$[\because \sin A \cos B-\sin B \cos A=\sin (A-B)$

$\& \sin A \cos B+\sin B \cos A=\sin (A+B)]$

$=\lim _{h \rightarrow 0} \frac{[\sin h][\sin (2 x+h)]}{h\left[\cos ^{2}(x+h) \cos ^{2} x\right]}$

$=\frac{1}{\cos ^{2} x} \lim _{h \rightarrow 0} \frac{\sin h}{h} \times \lim _{h \rightarrow 0} \sin (2 x+h) \times \lim _{h \rightarrow 0} \frac{1}{\cos ^{2}(x+h)}$

$=\frac{1}{\cos ^{2} x} \times(1) \times \lim _{h \rightarrow 0} \sin (2 x+h) \times \lim _{h \rightarrow 0} \frac{1}{\cos ^{2}(x+h)}$

$\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]$

Putting h = 0, we get

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

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

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

$[\because \sin 2 x=2 \sin x \cos x]$

$=2 \frac{\sin x}{\cos x} \times \sec ^{2} x\left[\because \frac{1}{\cos x}=\sec x\right]$

$=2 \tan x \sec ^{2} x$

$\left[\because \frac{\sin x}{\cos x}=\tan x\right]$

Hence, $f^{\prime}(x)=2 \tan x \sec ^{2} x$

 

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