# Product Differentiation

• Mar 27th 2009, 10:52 AM
Junesong
Product Differentiation
This is a homework problem on the Algebra of Derivatives section in my text. I'm not sure if I'm handling the $\displaystyle f(x^3)$ part correctly. Any suggestions (or if it's right, let me know) would help ease my mind greatly.

Problem
$\displaystyle f:R \rightarrow R$ is differentiable.

$\displaystyle g(x)=x^2 f(x^3)$

Prove $\displaystyle g(x)$ is differentiable and find the derivative.

Proof
We know that $\displaystyle f(x)$ is differentiable over the real numbers, and clearly, if $\displaystyle x$ is a real number, then $\displaystyle x^3$ is a real number, so $\displaystyle f(x^3)$ is differentiable. Let $\displaystyle h(x)=x^2$. Then $\displaystyle h(x)$ is differentiable, and by the Algebra of Derivatives, $\displaystyle g(x)$ must be differentiable.

It follows from the product rule that $\displaystyle g(x)=2xf(x^3)+x^2 f'(x^3)$.
• Mar 27th 2009, 10:59 AM
HallsofIvy
Quote:

Originally Posted by Junesong
This is a homework problem on the Algebra of Derivatives section in my text. I'm not sure if I'm handling the $\displaystyle f(x^3)$ part correctly. Any suggestions (or if it's right, let me know) would help ease my mind greatly.

Problem
$\displaystyle f:R \rightarrow R$ is differentiable.

$\displaystyle g(x)=x^2 f(x^3)$

Prove $\displaystyle g(x)$ is differentiable and find the derivative.

Proof
We know that $\displaystyle f(x)$ is differentiable over the real numbers, and clearly, if $\displaystyle x$ is a real number, then $\displaystyle x^3$ is a real number, so $\displaystyle f(x^3)$ is differentiable. Let $\displaystyle h(x)=x^2$. Then $\displaystyle h(x)$ is differentiable, and by the Algebra of Derivatives, $\displaystyle g(x)$ must be differentiable.

It follows from the product rule that $\displaystyle g(x)=2xf(x^3)+x^2 f'(x^3)$.

You have differentiated $\displaystyle f(x^3)$ incorrectly. By the chain rule $\displaystyle \frac{df(x^3)}{dx}= f'(x^3) \frac{d x^3}{dx}= 3x^2f'(x^3)$.
That is what you must multiply by $\displaystyle x^2$.
• Mar 28th 2009, 08:11 AM
Junesong
Quote:

Originally Posted by HallsofIvy
You have differentiated $\displaystyle f(x^3)$ incorrectly. By the chain rule $\displaystyle \frac{df(x^3)}{dx}= f'(x^3) \frac{d x^3}{dx}= 3x^2f'(x^3)$.
That is what you must multiply by $\displaystyle x^2$.

Then $\displaystyle g(x)=2xf(x^3)+3x^4 f'(x^3)$?
• Mar 28th 2009, 09:32 AM
HallsofIvy
Yes.