Implicit Differentiation, and the Chain Rule

[QuarkCharmer]

Homework Statement

Use implicit differentiation to find dy/dx
2x^3+x^2y-xy^3 = 2

Homework Equations

Chain Rule et al.

The Attempt at a Solution

My questions is this. When deriving something like xy^3, apply the product rule to get
1y^3 + x\frac{d}{dx}y^3

I am confused on differentiating y^3. Is it
3y^2y'
??

Making the solution to the mentioned equation:

y' = \frac{6x^2+2xy-y^3}{3xy^2-x^2}

 

[Mark44]

Homework Statement

Use implicit differentiation to find dy/dx
2x^3+x^2y-xy^3 = 2

Homework Equations

Chain Rule et al.

The Attempt at a Solution

My questions is this. When deriving something like xy^3, apply the product rule to get
1y^3 + x\frac{d}{dx}y^3

I am confused on differentiating y^3. Is it
3y^2y'
??

Yes.

Making the solution to the mentioned equation:

y' = \frac{6x^2+2xy-y^3}{3xy^2-x^2}

The rest is pretty much just algebra.

 

[QuarkCharmer]

Okay, so I am bringing down the exponent, applying the n-1, then multiplying it by the derivative of the thing in the x^n (the x). I get confused because when this happens with trigonometric equations it seems different. For instance:
sin^3(sin(x^3))

The derivative is then:
3sin^2(sinx^3)(3x^2) ?

I am sure there should be a cosine in there somewhere.

I want to say it SHOULD be:
3sin^2(sinx^3)cos(sinx^3)cos(x^3)3x^2

And as always, thank you.

 

[Mark44]

Okay, so I am bringing down the exponent, applying the n-1, then multiplying it by the derivative of the thing in the x^n (the x). I get confused because when this happens with trigonometric equations it seems different. For instance:
sin^3(sin(x^3))

The derivative is then:
3sin^2(sinx^3)(3x^2) ?

No.

I am sure there should be a cosine in there somewhere.

Yes.
You should get
3sin^2(sin(x^3))cos(x^3)(3x^2)

That cosine factor I added is the derivative of sin(x³) with respect to x³. The 3x² factor at the end is the derivative of x³ with respect to x.

I want to say it SHOULD be:
3sin^2(sinx^3)cos(sinx^3)cos(x^3)3x^2

And as always, thank you.