1. ## Second derivative help

Just wanted to see if I'm doing this correctly:

Find $f''(x)=|x|^3$

So $f'(x)=(3|x|^2)\frac{x}{|x|}$

$=3|x|x$

$f''(x)=3|x|\frac{d}{dx}x+3\frac{d}{dx}|x|x$

$=3|x|+3\frac{x}{|x|}x$

$=3|x|+\frac{3x^2}{|x|}$
Does that seem right?

2. Originally Posted by ascendancy523
Just wanted to see if I'm doing this correctly:

Find $f''(x)=|x|^3$

So $f'(x)=(3|x|^2)\frac{x}{|x|}$

$=3|x|x$

$f''(x)=3|x|\frac{d}{dx}x+3\frac{d}{dx}|x|x$

$=3|x|+3\frac{x}{|x|}x$

$=3|x|+\frac{3x^2}{|x|}$
Does that seem right?

First of all your first line is your second derivative... you need to ask : What is the second derivative of $f(x)$

Hint:

$f(x)=|x|^3=sgn(x)\cdot x^3$

Now, is much better looking...

You can see straight that:

$f''(x)=6\sqrt{x^2}$

3. Originally Posted by ascendancy523
Just wanted to see if I'm doing this correctly:

Find $f''(x)=|x|^3$
This is written badly. What this says is that $f''(x)$ is $|x|^3$. What you mean is "Find f''(x) where $f(x)= |x|^3$

So $f'(x)=(3|x|^2)\frac{x}{|x|}$

$=3|x|x$

$f''(x)=3|x|\frac{d}{dx}x+3\frac{d}{dx}|x|x$

$=3|x|+3\frac{x}{|x|}x$

$=3|x|+\frac{3x^2}{|x|}$
Does that seem right?
Of course, $x^2= |x|^2$
so this is the same as $3|x|+ 3\frac{|x|^2}{|x|}= 6|x|$ for $x\ne 0$ and is not defined for x= 0.

I would consider it simpler (and better) to say "If $x\ge 0$ $f(x)= x^3$ so $f'(x)= 3x^2$ and $f"(x)= 6x$, if x< 0, $f(x)= -x^3$ so $f'(x)= -3x^2$ and $f''(x)= -6x$. Of course, |x| is not differentiable at x= 0 so neither is $|x|^3$ and so f'' does not exist at x= 0".

4. Originally Posted by HallsofIvy
This is written badly. What this says is that $f''(x)$ is $|x|^3$. What you mean is "Find f''(x) where $f(x)= |x|^3$

Of course, $x^2= |x|^2$
so this is the same as $3|x|+ 3\frac{|x|^2}{|x|}= 6|x|$ for $x\ne 0$ and is not defined for x= 0.

I would consider it simpler (and better) to say "If $x\ge 0$ $f(x)= x^3$ so $f'(x)= 3x^2$ and $f"(x)= 6x$, if x< 0, $f(x)= -x^3$ so $f'(x)= -3x^2$ and $f''(x)= -6x$. Of course, |x| is not differentiable at x= 0 so neither is $|x|^3$ and so f'' does not exist at x= 0".
Yeah that would make it much simpler. Thanks for the help!