# Homework Help: Derivative of cubic absolute value function

1. Dec 7, 2011

### aero_zeppelin

1. The problem statement, all variables and given/known data

f(x) = |x+2|^3 -1

3. The attempt at a solution

- So, I know the formula is: d|x| / dx = x / |x|

- My guess would be: (x+2)^3 / |x + 2|^3 ???

I'm trying to find Increase and Decrease intervals for the graph..

2. Dec 7, 2011

### eumyang

No. Looks like you'll need to use the chain rule.

3. Dec 7, 2011

### aero_zeppelin

ok... hmmm , so maybe like this? :

(3) |x+2|^2 ( x+2 / |x+2| )

4. Dec 7, 2011

### Char. Limit

That sounds right.

5. Dec 7, 2011

### aero_zeppelin

Great, thanks... Now question #2:

I'm testing for concavity, so I'm getting the 2nd derivative...

For the x+2 / |x+2| part, does the quotient rule apply as normal??

The procedure would be a chain rule involving a product rule and a quotient rule right?

6. Dec 7, 2011

### Char. Limit

It might help if you simplified a bit, getting rid of an |x+2| in the numerator and denominator. Then you just have 3 (x+2) |x+2|, and that's simple to solve with the product rule.

7. Dec 7, 2011

### aero_zeppelin

You're right. So I guess the second derivative would be:

f '' (x) = ( 3 |x+2| ) + 3 (x+2) (x+2) / |x+2|

Thanks a lot btw

8. Dec 7, 2011

### SammyS

Staff Emeritus
To find the derivative of $\displaystyle \frac{x+2}{|x+2|}\,,$ simply consider its graph. The derivative is undefined at x=-2 and is 1 everywhere else.

9. Dec 7, 2011

### SammyS

Staff Emeritus
In my opinion, the clearest way to do this is with the piecewise definition of |x + 2|.

$\displaystyle f(x) = \left|x+2\right|^3=\left\{ \matrix{(x+2)^3,\ \ \text{if}\quad x>-2 \\ \ \ 0\ ,\ \ \quad\quad\quad\text{if}\quad x=-2 \\ -(x+2)^3,\ \ \text{if}\quad x<-2} \right.$

Finding the first & second derivatives is straight forward, if x ≠ -2. Take the limits from the left & right to show that f ' (-2) = 0 and f '' (-2) = 0

10. Dec 7, 2011

### Char. Limit

Considering that OP started with this as the expression for the derivative:

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

It'd probably just be easier to cancel out an |x+2| term from the numerator and denominator. It's easy to find the second derivative then.