Question

Find dy/dx of x+sin(x+y)=y-cos(x-y)

Answer 1

Step-by-step explanation:

The value of dy/dx is solved in the picture

Answer 2

Given :

x + sin(x+y) = y - cos(x-y)

To Find :

Derivative of x+sin(x+y)=y-cos(x-y)

Step-by-step explanation:

$\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'$

$=\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\sin \left(x+y\right)\right)-\frac{d}{dx}\left(y\right)+\frac{d}{dx}\left(\cos \left(x-y\right)\right)$

(1) $\frac{d}{dx}\left(x\right) = 1$

(2) $\frac{d}{dx}\left(\sin \left(x+y\right)\right) = cos(x+y) + cos(x+y)(\frac{dy}{dx} )$

(3) $-\frac{d}{dx}\left(y\right) = -\frac{dy}{dx}$

(4) $\frac{d}{dx}\left(\cos \left(x-y\right)\right)$

Adding up all , we get ;

=> 0 = $1+\cos \left(x+y\right)\left(1+\frac{d}{dx}\left(y\right)\right)-\frac{d}{dx}\left(y\right)-\sin \left(x-y\right)\left(1-\frac{d}{dx}\left(y\right)\right)$

taking $\frac{dy}{dx}$ on the left hand side of the equation , we get :

$\frac{dy}{dx}$  $= \frac{1+\cos \left(x+y\right)\left -\sin \left(x-y\right)\left }{-\cos \left(x+y\right)\left +1 -\sin \left(x-y\right)\left}$

Hence , the derivative of the given equation is :

$\frac{1+\cos \left(x+y\right)\left -\sin \left(x-y\right)\left }{1-\cos \left(x+y\right)\left -\sin \left(x-y\right)\left}$

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