The derivative of ^-1( x- x x+ x), with respect to x2, where x(0,2), is:

Question

The derivative of $	an^{-1}\left(\dfrac{\sin x-\cos x}{\sin x+\cos x}ight)$, with respect to $\dfrac x2$, where $x\in\left(0,\dfrac\pi2ight)$, is:

Accepted Answer

$$y = {\tan ^{ - 1}}\left( {{{\tan x - 1} \over {\tan x + 1}}} \right)$$

$$ \Rightarrow $$ $$ - {\tan ^{ - 1}}\left( {{{1 - \tan x} \over {1 + \tan x}}} \right)$$

$$ \Rightarrow $$ $$ - {\tan ^{ - 1}}\left( {\tan \left( {{\pi  \over 4} - x} \right)} \right)$$

$$ \Rightarrow $$ $$ - \left( {{\pi  \over 4} - x} \right)$$

$$ \Rightarrow $$ $${{dy} \over {dx}} = 1$$

Now if differentiation of $${x \over 2}$$ w.r.t is $${1 \over 2}$$

$$ \Rightarrow $$ So differentiation of y w.r.t $${x \over 2}$$ is $$1 \over {1 \over 2}$$ = 2

Solutions · 1