So the correct answer isdy/dx = (-sin4x - 4siny) / (4y + 4cos4x)

[JackieAnne]

Use implicit differentiation to find dy/dx.
y is a differentiable function of x

2y^2+4xsiny = cos4x

Here is what I did:

4y*dy/dx + 4siny+ 1/cosy*dy/dx = -sin4x + 4
4y*dy/dx + 1/cosy*dy/dx = -sin4x - 4siny + 4
dy/dx(4y + 1/cosy) = -sin4x - 4siny + 4
dy/dx = (-sin4x - 4siny + 4) / (4y + (1/cosy))

My answer is incorrect. Can anyone tell me where I went wrong and what the answer should be? Thanks

 

[fzero]

2y^2+4xsiny = cos4x

Here is what I did:

4y*dy/dx + 4siny+ 1/cosy*dy/dx = -sin4x + 4

For the bold part on the left hand side, you want to use

\frac{d}{dy} \sin y = \cos y,

not 1/cos y. For the bold part on the right hand side, use the chain rule to see that

\frac{d}{dx} \cos(4x) = - \sin(4x) \frac{d(4x)}{dx} = - 4\sin(4x) .