Question

find dy/dx if x cosy +y cosx =0

Answer 1

Solution -
-xsinydy/dx+cosy-ysinx+cosxdy/dx=0
dy/dx(cosx-xsiny)=ysinx-cosy
dy/dx=(ysinx-cosy)/(cosx-xsiny)

Answer 2

Answer:

$\frac{dy}{dx}=\frac{y\sin x+\cos y}{-x\sin y+\cos x}$

Step-by-step explanation:

The given equation is $x\cos y+y\cos x=0$

Differentiating both sides with respect to x. Apply the product rule of differentiation

$x(-\sin y)\frac{dy}{dx}+\cos y+y(-\sin x)+\cos x\cdot\frac{dy}{dx}$

Take dy/dx common, we get

$(-x\sin y+\cos x)\frac{dy}{dx}+\cos y-y\sin x=0$

Take cosy-ysinx to other side of the equation

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

Divide both sides by -xsiny+cosx

$\frac{dy}{dx}=\frac{y\sin x-\cos y}{-x\sin y+\cos x}$