Derivative with Absolute Value

[science.girl]

Homework Statement

Find f'(x), if

f(x) = [x^2 * (3x + 2)^(1/3)] / [(2x - 3)³]

Where the absolute value symbol surrounds the entire function.

Homework Equations

N/A

The Attempt at a Solution

My attempts don't account for the absolute value of the function. Otherwise, I can still take the derivative using quotient rule, product rule, and chain rule. Must I approach things differently because of the absolute value symbols? If so, how?

 

[Mark44]

Is this your function?
\frac{|x^2(3x + 2)^{1/3}|}{(2x - 3)^3}

If so, you can simplify it a bit to this:
\frac{x^2|(3x + 2)^{1/3}|}{(2x - 3)^3}

You will need to take into account the absolute values of the factor in the numerator and the one in the denominator. One way to do this is to look at your function on three different intervals: -infinity < x <-2/3, -2/3 < x < 3/2, and 3/2 < x < infinity.
Each of these intervals gives a different version of your function if you remove the absolute values, so each will give you a different version of the derivative.

 

[arch]

Homework Statement

Find f'(x), if

f(x) = [x^2 * (3x + 2)^(1/3)] / [(2x - 3)³]
?

\frac{x^2}{(-3+2 x)^3 (2+3 x)^{2/3}}-\frac{6 x^2 (2+3 x)^{1/3}}{(-3+2 x)^4}+\frac{2 x (2+3 x)^{1/3}}{(-3+2 x)^3}

x^2/((-3 + 2 x)^3 (2 + 3 x)^(2/3)) - (
6 x^2 (2 + 3 x)^(1/3))/(-3 + 2 x)^4 + (
2 x (2 + 3 x)^(1/3))/(-3 + 2 x)^3

Homework Equations

N/A

?

The Attempt at a Solution

My attempts don't account for the absolute value of the function. Otherwise, I can still take the derivative using quotient rule, product rule, and chain rule. Must I approach things differently because of the absolute value symbols? If so, how?

 

[Bohrok]

You can also use the chain rule knowing that |x| = √(x²)

 

[science.girl]

Ok; I think I understand how to approach this now. I'll get more help from my instructor to be sure. (I completed the assignment before, but wanted to understand for the final.)

So, thanks!