Y cos x 2

Asked by maham237 @ in Mathematics viewed by 322 People

Implicit differentiation: ycosx=x^2+y^2
Derivative of both sides:
=((y)(-sinx)+(cosx)(dy/dx)=2x+2y(dy/dx)
=(-ysinx+cosx(dy/dx)=2x+2y(dy/dx)
=cosx(dy/dx) - 2y(dy/dx) = ysinx+2x
=(cosx-2y)dy/dx = ysinx+2x
=dy/dx = (ysinx+2x)/(cosx-2y)

Answered by maham237 @


Y cos x = x^2 + y^2

Taking derivative:

y' cos x - y sin x = 2x + 2y y'

Simplifying:

y' = (- 2x - y sin x)/(2y - cos x)

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