Question

Differentiate from first principle:
$\left(iv\right)tan{x}^2$

Answer 1

Given:

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

The derivative of a function $f\left(x\right)$ is defined as:

$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$

Putting $f\left(x\right)$ in above expression, we get:
$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{tan{(x+h)}^2-tan{x}^2}{h}$

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{sin{(x+h)}^{2}cos{x}^2-sin{x}^{2}cos{(x+h)}^2}{hcos{(x+h)}^{2}cos{x}^2}$

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{sin[{(x+h)}^2-x^2]}{hcos{(x+h)}^{2}cos{x}^2}$

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{sin(h^2+2hx)}{(h^2+2hx)}\times \frac{(h^2+2hx)}{h}\times \frac{1}{cos{(x+h)}^{2}cos{x}^2}$

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{sin(h^2+2hx)}{(h^2+2hx)}\times (h+2x)\times \frac{1}{cos{(x+h)}^{2}cos{x}^2}$

$\Rightarrow f^{\prime }\left(x\right)=1\times (0+2x)\times \frac{1}{cos{(x+0)}^{2}cos{x}^2}$

$[\because \underset{h\to 0}{lim}\frac{sin\left(h\right)}{h}=1]$

$\Rightarrow f^{\prime }\left(x\right)=2xse{c}^2{x}^2$

Therefore, the derivative of $tan{x}^2$ is $2xse{c}^2{x}^2$