Question

find derivative of (i)cot^-1(cosx-sinx)/(cosx+sinx)​

Answer 1

Given Expression,

$\sf \: f(x) = {cot}^{ - 1} \bigg( \dfrac{cos \: x - sin \: x}{cos \: x + sin \: x} \bigg)$

We know that,

$\star \: \boxed{ \boxed{ \sf {tan}^{ - 1} x = {cot}^{ - 1} \big( \dfrac{1}{x} \big)}}$

Dividing the numerator and denominator by sin x,

$\longrightarrow \: \sf \: f(x) = {tan}^{ - 1} \bigg( \dfrac{ \frac{cos \: x + sin \: x}{cos \: x}}{ \frac{cos \: x - sin \: x}{cos \: x}} \bigg) \\ \\ \longrightarrow \: \sf f(x) = {tan}^{ - 1} \bigg( \dfrac{tan( \frac{\pi}{4}) + tan \: x}{1 - tan \: x.tan( \frac{\pi}{4} )} \bigg)$

We know that,

$\star \: \boxed{ \boxed{ \sf {tan}^{}( x + y) = \frac{tan \: x + tan \: y}{1 - tan \: x.tan \: y} \big)}}$

Thus,

$\longrightarrow \sf \: f(x) = {tan}^{ - 1} \bigg( tan(x + \dfrac{\pi}{4}) \bigg) \\ \\ \longrightarrow \sf \: f(x) = x + \dfrac{\pi}{4} \\ \\ \longrightarrow \boxed{\boxed{ \sf \: f'(x) = 1}}$