Product Differentiation

**Junesong** · March 27th 2009, 10:52 AM

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

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

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

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

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

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

**HallsofIvy** · March 27th 2009, 10:59 AM

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 $f(x^3)$ part correctly. Any suggestions (or if it's right, let me know) would help ease my mind greatly.

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

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

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

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

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

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

**Junesong** · March 28th 2009, 08:11 AM

Originally Posted by HallsofIvy

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

Then $g(x)=2xf(x^3)+3x^4 f'(x^3)$ ?

**HallsofIvy** · March 28th 2009, 09:32 AM

Yes.

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