Question

Derivative of √tan^-1 (x/2)

Answer 1

$\huge{\bold{\underline{\underline{Question:-}}}}$

Derivative of ${tan}^{-1) \times (\frac{x}{2})$

$\huge{\bold{\underline{\underline{Solution:-}}}}$

$\frac{d}{dx}({tan}^{-1}) = \frac{1}{1+{x}^{2}}$

Differentiate using the chain rule

Given f(x) = g(h(x)) then

f ' (x) = g ' (h(x)) × h ' (x) ⪻ Chain rule

$f(x) = {tan}^{-1) \times (\frac{x}{2})$

$\implies{f'(x)\:=\:\frac{1}{1+{\frac{x}{2}}^{2} \times [tex]\frac{d}{dx} \times ({x \frac{1}{2})$

$=\frac{\frac{1}{2}}{1+{\frac{{x}^{2}}{2}} \times \frac{4}{4}$

$= \frac{2}{4+{x}^{2}}$