Solving Derivative Problem: Find f'(x)

[camboguy]

Homework Equations

find
f'(x)= [(x^2)(x-3)]^(1/3)

The Attempt at a Solution

i got (3(x-2))/[x(x-3)^(2/3)]

is that correct? I am so confused if i got it right or not.

Thanks before hand

 

[Lazerlike42]

Not quite. I'd start by making it simpler by multiplying out the polynomials. Then you'd have:

(x^{3}-3x^{2})^{\frac{1}{3}}

Then, use the chain rule:

If f(x)=g(h(x)), then f ' (x) = g ' (f(x))*f ' (x).

For example, look at f(x) = (sin(x))².

So if g(x) = x², and h(x) = sin(x), then f(x)=g(h(x)). Then to take the derivative, first, find the derivative of g and h.

So g(x)=x², which means that g ' (x)=2x.
h(x) = sin(x), which means that h ' (x)= cos(x).

Then to take the derivative, you would remember that f ' (x) = g ' (f(x))*f ' (x). So first, look at g ' (x). That's 2x. So put f(x) in for x, to get 2(sin(x)). Then, multiply by f ' (x), which is cos(x).

So f ' (x), in this example, is 2(sin(x))*cos(x).

Now, use that same method for your problem.

 

[CalculusSandwich]

This is an application of the chain rule.. You can multiply the inner product and get...[x^3-3x^2]

... now you bring the 1/3 out in front and reduce it by 1 as by application of the derivative. 1/3-1/1 = -2/3

So you get what 1/3[d/dx[x^3-3x^2]]^-2/3

so then it's: 1/3[x^2*(x-3)]^(-2/3) * derivative of the inside [3x^2-6x]

Got it?

 

[Mentallic]

never mind -.-

 

[Mark44]

Homework Equations

find
f'(x)= [(x^2)(x-3)]^(1/3)

If you're trying to find f'(x), look no further than the equation you wrote.

I'm being facetious here. What you most likely meant was:
f(x)= [(x^2)(x-3)]^(1/3)
Find f'(x).

This might seem like a small difference, but it's important to keep a function and its derivatives clearly separated in your mind.

The Attempt at a Solution

i got (3(x-2))/[x(x-3)^(2/3)]

is that correct? I am so confused if i got it right or not.

Thanks before hand