Use this equation to find dy/dx. 8y cos(x) = x² + y²

Solution:

Given 8ycos(x) = x² + y²

This is an implicit function

By implicit differentiation, we get

⇒ d/dx {8ycosx} = d/dx(x² + y²)

⇒ 8[dy/dxcosx + y(-sinx)] = 2x + 2ydy/dx

⇒8[dy/dxcosx + y(-sinx)] = 2[x + ydy/dx]

⇒4[dy/dxcosx + y(-sinx)] = [x + ydy/dx]

⇒4dy/dxcosx + -4y sinx = x + ydy/dx

⇒4dy/dx. cosx -ydy/dx = x + 4y sinx

⇒dy/dx(4 cos x -y)= x + 4y sinx

⇒ dy/dx= [x + 4y sinx]/[(4 cos x -y)]

Use this equation to find dy/dx. 8y cos(x) = x² + y²

Summary:

The value of dy/dx of 8y cos(x) = x² + y² is [x + 4y sinx]/[(4 cos x -y)]