Finding f'(x) for Logarithm Derivatives Homework with Product and Quotient Rules

[Justabeginner]

Homework Statement

Find f'(x): $(abs(((x^2)*((3x+2) ^(1/3)))/((2x-3)^3))$ <- Not sure why it''s not showing up but the 1/3 is an exponent to just the (3x+2).

Homework Equations

Product Rule and Quotient Rule for Differentiating

The Attempt at a Solution

So I thought I should split it into two parts: A and B.

A) $(x^2)/((2x-3)^2)$
B) $((3x+2)^1/3)/(2x-3)$

My plan was to differentiate each part and then multiply them together, since multiplying their denominators together gives me the original denominator power (4).

I end up getting for A) (-12x^2 + 18x)/((2x-3)^4)

For B, I have something which seems much more complex, because of the negative fractional exponents, and this is where I'm stuck. I want to simplify, but I'm not sure how to without screwing everything up. This is my last step for B:

$((3x+2)^(-2/3)*(2x-3) - 4(x-3)*((3x+2)^(1/3)))/((2x-3)^2)$ <- The -2/3 is the exponent

I really appreciate the help. And also, as for the absolute value part of the question, well I haven't even thought about that- I'm stuck on the easy part itself. -_-

Thank you!

 

[Mark44]

Homework Statement

Find f'(x): $(abs(((x^2)*((3x+2) ^(1/3)))/((2x-3)^3))$ <- Not sure why it''s not showing up but the 1/3 is an exponent to just the (3x+2).

In LaTeX, if an exponent is more than one character, you have to put braces around the entire expression of the exponent. If your exponent expression includes parentheses, they need to go inside the braces as well.

So (3x + 2)^1/3 would be written as (3x + 2)^{1/3}.

Homework Equations

Product Rule and Quotient Rule for Differentiating

The Attempt at a Solution

So I thought I should split it into two parts: A and B.

No, you can't do this. It's not true that d/dx( f(x) * g(x)) = f'(x) * g'(x). That seems to be what you're trying to do, from what you're saying below.

A) $(x^2)/((2x-3)^2)$
B) $((3x+2)^1/3)/(2x-3)$

My plan was to differentiate each part and then multiply them together, since multiplying their denominators together gives me the original denominator power (4).

I end up getting for A) (-12x^2 + 18x)/((2x-3)^4)

For B, I have something which seems much more complex, because of the negative fractional exponents, and this is where I'm stuck. I want to simplify, but I'm not sure how to without screwing everything up. This is my last step for B:

$((3x+2)^(-2/3)*(2x-3) - 4(x-3)*((3x+2)^(1/3)))/((2x-3)^2)$ <- The -2/3 is the exponent

I really appreciate the help. And also, as for the absolute value part of the question, well I haven't even thought about that- I'm stuck on the easy part itself. -_-

d/dx |x| = x/|x|, and
d/dx |u| = u/|u| * du/dx

Thank you!

 

[Justabeginner]

Oh wow, that was the rule that we learned in school- I should have known I'm not supposed to do that! I'll redo the problem, and see what I get. Thanks!

 

[Justabeginner]

If $d/dx [x] = x/|x|$ ,

then would the correct expression be $d/dx |x| = (f(x))/|((x^2)((3x+2)^{1/3}))/((2x-3)^3)|$ ?

meaning, in simplified form would it be: $((f(x)) * ((2x-3)^3)/((x^2)(3x+2)^{1/3})$

Thank you.