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.