Question

Differentiate from first principle:
$\left(ii\right)tan(2x+1)$

Answer 1

$f\left(x\right)=tan(2x+1)$

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:

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{tan[2(x+h)+1]-tan(2x+1)}{h}$

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{\begin{array}{c}sin(2x+2h+1)cos(2x+1)-cos(2x+2h+1)sin(2x+1)\end{array}}{hcos(2x+2h+1)cos(2x+1)}$

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

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

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

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

$\Rightarrow f^{\prime }\left(x\right)=\frac{2}{co{s}^2(2x+1)}$

$\Rightarrow f^{\prime }\left(x\right)=2se{c}^2(2x+1)$