Product Rule for Derivatives: Understanding and Applying

[Winning]

Homework Statement

$$\frac{d}{dx}x^{2}y^{2} = ?$$

Homework Equations

Power rule: $$\frac{d}{dx}x^{n} = nx^{n-1}$$
Product rule: $$\frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + g'(x)f(x)$$

The Attempt at a Solution

My TI-89 titanium gives me $$2xy^{2}$$
Which contradicts the power rule.

Doing this by hand gives me
$$2x^{2}y + 2xy^{2}$$

Which one is correct? Is "y" considered as it's own function or as a constant?

 

[Ray Vickson]

Homework Statement

$$\frac{d}{dx}x^{2}y^{2} = ?$$

Homework Equations

Power rule: $$\frac{d}{dx}x^{n} = nx^{n-1}$$
Product rule: $$\frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + g'(x)f(x)$$

The Attempt at a Solution

My TI-89 titanium gives me $$2xy^{2}$$
Which contradicts the power rule.

Doing this by hand gives me
$$2x^{2}y + 2xy^{2}$$

Which one is correct? Is "y" considered as it's own function or as a constant?

Why do you think that d(y^2)/dx is 2y?

RGV

 

[jbunniii]

If y does not depend on x, then the calculator's answer is correct. You treat y as a constant in that case.

If y does depend on x, then you have to use the chain rule in addition to the rules you used:

$$\frac{d}{dx} (x^2 y^2) = 2xy^2 + 2x^2 y \frac{dy}{dx}$$

Notice that if y is constant (not dependent on x), then

$$\frac{dy}{dx} = 0$$

so the above reduces to your calculator's answer.

 

[Winning]

@Ray Vickson: That's a good question... I had a brainfart lol.

@jbunniii: Yeah, y is dependent on x... I forgot to mention that this is part of an Imp. Diff. problem. Thanks!