Question

if y=sin(x+a)/cosx then find dy/dx​

Answer 1

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

Given that

$\rm :\longmapsto\:y = \dfrac{sin(x + a)}{cosx}$

We know,

$\boxed{ \bf{ \: sin(x + y) = sinxcosy + sinycosx}}$

Using this identity we get,

$\rm :\longmapsto\:y = \dfrac{sinx \: cosa + sina \: cosx}{cosx}$

$\rm :\longmapsto\:y = \dfrac{sinx \: cosa }{cosx} \: + \: \dfrac{sina \: cosx}{cosx}$

$\rm :\longmapsto\:y \: = \: cosa \: tanx \: + \: sina$

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

$\rm :\longmapsto\:\dfrac{d}{dx}y \: = \dfrac{d}{dx}( \: cosa \: tanx \: + \: sina)$

$\rm :\longmapsto\:\dfrac{dy}{dx} \: = \dfrac{d}{dx} \: cosa \: tanx \: + \dfrac{d}{dx} \: sina$

We know,

$\boxed{ \bf{ \: \dfrac{d}{dx}k = 0}}$

and

$\boxed{ \bf{ \: \dfrac{d}{dx}k \: f(x)= k \: \dfrac{d}{dx} \: f(x)}}$

So, using these, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = cosa \: \dfrac{d}{dx} \: tanx \: + \: 0$

$\bf :\longmapsto\:\dfrac{dy}{dx} = cosa \: {sec}^{2} x$

Alternative Method :-

Given that

$\rm :\longmapsto\:y = \dfrac{sin(x + a)}{cosx}$

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

$\rm :\longmapsto\:\dfrac{d}{dx}y =\dfrac{d}{dx} \dfrac{sin(x + a)}{cosx}$

We know,

$\boxed{ \bf{ \: \dfrac{d}{dx} \frac{u}{v} = \frac{v\dfrac{d}{dx}u \: - \: u\dfrac{d}{dx}v}{ {v}^{2} } }}$

So, using this we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{cosx\dfrac{d}{dx}sin(x + a) - sin(x + a)\dfrac{d}{dx}cosx}{ {cos}^{2}x }$

We know,

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

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

Using these, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{cosx \: cos(x + a) - sin(x + a) \: ( - sinx)}{ {cos}^{2}x }$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{cosx \: cos(x + a) + sin(x + a) \: sinx}{ {cos}^{2}x }$

We know,

$\boxed{ \bf{ \: cosx \: cosy \: + \: sinx \: siny \: = \: cos(x - y)}}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{cos(x + a - x)}{ {cos}^{2} x}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{cosa}{ {cos}^{2} x}$

$\bf :\longmapsto\:\dfrac{dy}{dx} = cosa \: {sec}^{2} x$

Additional Information :-

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

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

$\boxed{ \bf{ \: \dfrac{d}{dx}cotx = - \: {cosec} \: cotx}}$

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

Answer 2

Step-by-step explanation:

Before differenciate it, we have to know some formula .

$\pink{\sf \: \odot \: \frac{d \: {x}^{n} }{dx} = n {x}^{n - 1} } \\ \\\orange{ \sf \: \odot \: \frac{d \: sin \: x}{dx} = cos \: x} \\ \\ \red{ \sf \: \odot \: \frac{d \: cos \: x}{dx} = \: - sin \: x }\\ \\ \green{\sf \: \odot \: \frac{d}{dx} ( \frac{u}{v} ) = \frac{v \frac{d \: u}{dx} - u \frac{d \: v}{dx} }{ {v}^{2} }} \\ \\\sf \: \odot \: \frac{d \: (constant) }{dx} = 0\\\\ \sf \: \odot \: cos(x - y) = cos \: x.cos \: y + sin \: x.sin \: y \\ \\ \sf \: \odot \: cos( - \theta) = cos \theta$

Now solution :

Given,

$\: \: \: \sf \: y = \frac{ \: sin(x + a)}{cos \: x} \\ \\ \sf \implies \frac{dy}{dx} = \frac{d}{dx} ( \frac{sin(x + a)}{cos \: x} ) \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{cos \: x \frac{d sin(x + a)}{dx} - sin(x + a) \frac{d \: cos \: x}{dx} }{ {cos}^{2} x} \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{cos \: x.cos(x + a) \frac{d(x + a)}{dx} - sin(x + a)( - sin \: x) }{ {cos}^{2}x } \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{cos \: x.cos(x + a) (1 + 0){}{} - sin(x + a)( - sin \: x) }{ {cos}^{2}x } \: \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{cos \: x.cos(x + a) + sin(x + a)( sin \: x) }{ {cos}^{2}x } \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{cos(x - x - a)}{ {cos}^{2}x } \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{cos( - a)}{ {cos}^{2} x} \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{cos \: a}{ {cos}^{2} x}\\\\\sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:=cos \:a. {sec}^2\:x$