Derivative of cubic absolute value function

[aero_zeppelin]

Homework Statement

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

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..

 

[eumyang]

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

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

 

[aero_zeppelin]

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

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

 

[Char. Limit]

That sounds right.

 

[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?

 

[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.

 

[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

 

[SammyS]

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?

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.

 

[SammyS]

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?

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\{<br /> \matrix{(x+2)^3,\ \ \text{if}\quad x&gt;-2 \\<br /> \ \ 0\ ,\ \ \quad\quad\quad\text{if}\quad x=-2 \\<br /> -(x+2)^3,\ \ \text{if}\quad x&lt;-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

 

[Char. Limit]

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\{<br /> \matrix{(x+2)^3,\ \ \text{if}\quad x&gt;-2 \\<br /> \ \ 0\ ,\ \ \quad\quad\quad\text{if}\quad x=-2 \\<br /> -(x+2)^3,\ \ \text{if}\quad x&lt;-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

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

f&#039;(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.