Question

differentiate with respect to x using chain rule​

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\: {tan}^{ - 1}\bigg(\dfrac{x - a}{1 + xa} \bigg)$

Let assume that,

$\rm :\longmapsto\: y \: = \: {tan}^{ - 1}\bigg(\dfrac{x - a}{1 + xa} \bigg)$

We know that,

$\green{\boxed{ \bf{ {tan}^{ - 1}x - {tan}^{ - 1}y = {tan}^{ - 1}\bigg(\dfrac{x - y}{1 + xy} \bigg) }}}$

So, using this identity we get

$\rm :\longmapsto\:y = {tan}^{ - 1}x - {tan}^{ - 1}a$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}y =\dfrac{d}{dx}( {tan}^{ - 1}x - {tan}^{ - 1}a)$

We know,

$\green{\boxed{ \bf{ \dfrac{d}{dx}(u + v) = \dfrac{d}{dx}u+ \dfrac{d}{dx}v}}}$

So, using this we get

$\rm :\longmapsto\:\dfrac{dy}{dx} =\dfrac{d}{dx}{tan}^{ - 1}x - \dfrac{d}{dx}{tan}^{ - 1}a$

We know that,

$\boxed{ \rm{ \dfrac{d}{dx}{tan}^{ - 1}x = \frac{1}{1 + {x}^{2} } }}$

and

$\boxed{ \rm{ \dfrac{d}{dx}k = 0}}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{1 + {x}^{2} } - 0$

$\bf :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{1 + {x}^{2} }$

Additional Information :-

$\boxed{ \rm{ \dfrac{d}{dx}{sin}^{ - 1}x = \frac{1}{ \sqrt{1 - {x}^{2} } } }}$

$\boxed{ \rm{ \dfrac{d}{dx}{cos}^{ - 1}x = \frac{ - \: 1}{ \sqrt{1 - {x}^{2} } } }}$

$\boxed{ \rm{ \dfrac{d}{dx}{cosec}^{ - 1}x = \frac{ - \: 1}{x \sqrt{ {x}^{2} - 1 } } }}$

$\boxed{ \rm{ \dfrac{d}{dx}{sec}^{ - 1}x = \frac{ \: 1}{x \sqrt{ {x}^{2} - 1 } } }}$

$\boxed{ \rm{ \dfrac{d}{dx}{cot}^{ - 1}x = \frac{ - \: 1}{ {x}^{2} + 1} }}$

$\boxed{ \rm{ \dfrac{d}{dx}sinx = cosx}}$

$\boxed{ \rm{ \dfrac{d}{dx}cosx = - sinx}}$

$\boxed{ \rm{ \dfrac{d}{dx}tanx = {sec}^{2} x}}$

$\boxed{ \rm{ \dfrac{d}{dx}cotx = { - \: cosec}^{2} x}}$