Question

Differentiate from first principle:
$\left(iv\right)\sqrt{tanx}$

Answer 1

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

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:

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

Rationalizing the numerator, we get:
$\Rightarrow f^{\prime }\left(x\right)$

$=\underset{h\to 0}{lim}\frac{\sqrt{tan(x+h)}-\sqrt{tanx}}{h}\times \frac{\sqrt{tan(x+h)}+\sqrt{tanx}}{\sqrt{tan(x+h)}+\sqrt{tanx}}$

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

$\Rightarrow f^{\prime }\left(x\right)$

$=\underset{h\to 0}{lim}\frac{sin(x+h)cosx-cos(x+h)sinx}{(h\sqrt{tan(x+h)}+\sqrt{tanx})cos(x+h)cosx}$

$\Rightarrow f^{\prime }\left(x\right)$

$=\underset{h\to 0}{lim}\frac{sin(x+h-x)}{h(\sqrt{tan(x+h)}+\sqrt{tanx})cos(x+h)cosx}$

$\Rightarrow f^{\prime }\left(x\right)$

$=\underset{h\to 0}{lim}\frac{sinh}{h(\sqrt{tan(x+h)}+\sqrt{tanx})cos(x+h)cosx}$

$\Rightarrow f^{\prime }\left(x\right)$

$=\underset{h\to 0}{lim}\frac{sinh}{h}\times \frac{1}{(\sqrt{tan(x+h)}+\sqrt{tanx})cos(x+h)cosx}$

$\Rightarrow f^{\prime }\left(x\right)$

$=1\times \frac{1}{(\sqrt{tan(x+0)}+\sqrt{tanx})cos(x+0)cosx}$

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

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