Small Confusion with Partial Derivative

[jegues]

Homework Statement

Let $$u(x,y) = f(x^3 + y^2) +g(x^3 + y^2)$$ such that f and g are differentiable functions. Show that,

$$2y\frac{\partial u}{\partial x} - 3x^{2} \frac{\partial u}{\partial y} = 0$$

Homework Equations

The Attempt at a Solution

The part of confused about is how to break down my partial derivatives.

The first thing I'm going to do is,

$$\text{Let } p=x^3 + y^2$$

then,

$$u = f(p) + g(p)$$

Now how to I extract,

$$\frac{\partial u}{\partial x},\frac{\partial u}{\partial y}$$

from here?

Is it simply,

$$\frac{\partial u}{\partial x} = \frac{du}{dp} \frac{\partial p}{\partial x}$$

The part that bothers me is that the du on the top is not a $$\partial$$.

Is this correct?

 

[Char. Limit]

Homework Statement

Let $$u(x,y) = f(x^3 + y^2) +g(x^3 + y^2)$$ such that f and g are differentiable functions. Show that,

$$2y\frac{\partial u}{\partial x} - 3x^{2} \frac{\partial u}{\partial y} = 0$$

Homework Equations

The Attempt at a Solution

The part of confused about is how to break down my partial derivatives.

The first thing I'm going to do is,

$$\text{Let } p=x^3 + y^2$$

then,

$$u = f(p) + g(p)$$

Now how to I extract,

$$\frac{\partial u}{\partial x},\frac{\partial u}{\partial y}$$

from here?

Is it simply,

$$\frac{\partial u}{\partial x} = \frac{du}{dp} \frac{\partial p}{\partial x}$$

The part that bothers me is that the du on the top is not a $$\partial$$.

Is this correct?

I do believe it is. You're actually dealing with a composite function, which is why that du isn't partial.

 

[jegues]

I do believe it is. You're actually dealing with a composite function, which is why that du isn't partial.

I think I'm confused because it says show that,

$$2y\frac{\partial u}{\partial x} - 3x^{2} \frac{\partial u}{\partial y} = 0$$

and in here I see,

$$\frac{\partial u}{\partial x} = \frac{du}{dp} \frac{\partial p}{\partial x}$$


Do the two d's cancel out and the two p's cancel out?