Question

Differentiate from first principle:
$\left(iii\right)tan\sqrt{x}$

Answer 1

Given:
$f\left(x\right)=tan\sqrt{x}$

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:

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

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

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

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

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

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

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

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

Therefore, the derivative of $tan\sqrt{x}$ is $\frac{se{c}^2\sqrt{x}}{2\sqrt{x}}$