Help with Calculus derivative
How do I take the first derivative of x^1/3(x-4)^2/3
I know it looks really easy at first, using chain rule, etc, but it gets ugly really fast and I don't know what to do. How on earth to I proceed to answering this question? I don't expect anyone to write out to whole solution (would be great though) but if I could get some assistance on how to proceed?
The point I get to, and then I have no clue what to do is this:
1/3x^-2/3(x-4)^2/3 + 2/3x^1/3(x-4)^-1/3
What to do next is confusing me. I probably did it wrong, or I am approaching the question in the wrong way. By the way, my teacher said this is an extremely difficult question, so don't hate aha.
April 8, 2014
4 Comments • Newest first
[quote=Copyrighted]If you use product rule, there is no chain rule (this is a simple product rule problem).
Let f(x)=(x^1/3)
Let g(x)=(x-4)^(2/3)
Product rule: f'(x)g(x)+g'(x)f(x)=dy/dx where y=f(x)g(x)[/quote]
Ah I see. I got to the last step, before simplification.
http://symbolab.com/solver/derivative-calculator/%5Cfrac%7Bd%7D%7Bdx%7D(x)%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D(x-4)%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D
How do I go from the second last step, to the last?
EDIT: To above, yeah, ahaha, I have to simplify it. That is the hard part.:/
You did it correctly and that's the answer. If you have to simplify, I feel bad for you son; I had 99 problems but mandatory simplification on exams wasn't one!
[quote=Copyrighted]y'(x) = (3 x-4)/(3 (x-4)^(1/3) x^(2/3))
assuming your function is y=(x^1/3)((x-4)^(2/3))
The function you gave is very ambiguous btw, use a few more parenthesis[/quote]
I believe that is the answer. Can you explain to me how you got that?
2+2=4. I Hope that helped.