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)

Question

Adam Burnett

5 May, 09:13

+3

Answers (1)

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Barton · Answer 1

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)

Implicit differentiation: ycosx=x^2+y^2Derivative 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)