Find the derivative of f(x) = (x-1)^(2/3) using the definition and then rules.

[PuddySporty]

Homework Statement

f(x) = (x-1) ^2/3
We were asked to find the derivative using the alternate limit form when c = 1 and then I was then I was trying to find the derivative using the multiplication rule (u dv + v du).

Homework Equations

f'(u*v) = u dv + v du
f'(xⁿ) = nx^n-1

The Attempt at a Solution

When I did this, I got (2(x-1)^-1/3)/3 which doesn't match the correct answer I got when I used the alternate limit form: 1/(x-1)^(2/3)
My question is if you can use the exponent rule on a polynomial and why it's not working out to be the same thing. Also, if you use the alternate limit form, does that mean it's the derivative for that c value only?

 

[Mark44]

Homework Statement

f(x) = (x-1) ^2/3
We were asked to find the derivative using the alternate limit form when c = 1 and then I was then I was trying to find the derivative using the multiplication rule (u dv + v du).

Homework Equations

f'(u*v) = u dv + v du
f'(xⁿ) = nx^n-1

Your equations here are incorrect. The frist one should be (f * g)'(x) = f'(x) g(x) + f(x) g'(x)
or
d/dx(u v) = u dv/dx + v du/dx

The second one should be d/dx(xⁿ) = nx^{n - 1}

The notation f' (something) does not mean "take the derivative of something," which is what you seem to think it means. Instead, it means "the derivative of f, evaluated at something."

The Attempt at a Solution

When I did this, I got (2(x-1)^-1/3)/3 which doesn't match the correct answer I got when I used the alternate limit form: 1/(x-1)^(2/3)

If f(x) = (x - 1)^{2/3, then f'(x) =} 1/(x-1)^(2/3) is incorrect. The first answer you showed is correct. Another way to write it is (2/3)(x - 1)^-1/3.

My question is if you can use the exponent rule on a polynomial and why it's not working out to be the same thing. Also, if you use the alternate limit form, does that mean it's the derivative for that c value only?