Question

differentiate w.r.t. x
please help me asap​

Answer 1

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

Let assume that

$\rm :\longmapsto\:y = {cot}^{ - 1}\bigg(\dfrac{cosx}{1 + sinx} \bigg)$

We know,

$\boxed{ \rm{ cosx = sin\bigg(\dfrac{\pi}{2} - x\bigg) }}$

and

$\boxed{ \rm{ sinx = cos\bigg(\dfrac{\pi}{2} - x\bigg) }}$

So, using these Identities, the given function reduces to

$\rm :\longmapsto\:y = {cot}^{ - 1}\bigg(\dfrac{sin\bigg(\dfrac{\pi}{2} - x\bigg)}{1 + cos\bigg(\dfrac{\pi}{2} - x\bigg)} \bigg)$

Let we assume that

$\red{\rm :\longmapsto\:\bigg(\dfrac{\pi}{2} - x\bigg) = 2z}$

So, above function can be rewritten as

$\rm :\longmapsto\:y = {cot}^{ - 1}\bigg(\dfrac{sin2z}{1 + cos2z} \bigg)$

We know,

$\boxed{ \rm{ sin2z = 2sinzcosz}}$

and

$\boxed{ \rm{ 1 + cos2z = {2cos}^{2}z}}$

So, using these Identities, we get

$\rm :\longmapsto\:y = {cot}^{ - 1}\bigg(\dfrac{2sinz \: cosz}{ {2cos}^{2} z} \bigg)$

$\rm :\longmapsto\:y = {cot}^{ - 1}\bigg(\dfrac{sinz}{ {cos}z} \bigg)$

$\rm :\longmapsto\:y = {cot}^{ - 1}\bigg(tanz \bigg)$

$\rm :\longmapsto\:y = \dfrac{\pi}{2} - {tan}^{ - 1}\bigg(tanz \bigg)$

$\rm :\longmapsto\:y = \dfrac{\pi}{2} - z$

On substituting the value of z, we get

$\rm :\longmapsto\:y = \dfrac{\pi}{2} - \bigg(\dfrac{\pi}{4} - \dfrac{x}{2} \bigg)$

$\rm :\longmapsto\:y = \dfrac{\pi}{2} - \dfrac{\pi}{4} + \dfrac{x}{2}$

$\rm :\longmapsto\:y = \dfrac{\pi}{4} + \dfrac{x}{2}$

Now, on differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx} y = \dfrac{d}{dx} \dfrac{\pi}{4} + \dfrac{d}{dx}\dfrac{x}{2}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{2}$

Additional Information :-

$\boxed{ \rm{ \dfrac{d}{dx}x = 1}}$

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

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

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

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

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

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

$\boxed{ \rm{ \dfrac{d}{dx}cosecx \: = \: - \: cosecx \: cotx}}$

$\boxed{ \rm{ \dfrac{d}{dx}secx \: = \: \: secx \: tanx}}$