I need to prove that when f(x)=x^(1/3), f'(x)=(1/3)x^(-2/3) using the differential equation --> (f(x+h)-f(x))/h It's very tough, I've already tried for 2 hours and gotten pretty much nowhere, so definitely +rep to anyone who can do it. :yup:
Yay calculus y = x^(1/3) dy/dx = lim (y(x+h) - y(x))/h = ( (x+h)^(1/3) - x^(1/3) )/h Now just multiply the top and bottom by: ((x+h)^(1/3))^2 + ((x+h)^(1/3)*x^(1/3)) + (x^(1/3))^2 ^ that is basically to get rid of the cube roots on the top of the equation, recall (a-b)^3 = (a-b)(a^2 + ab + b^2) so in this case: a = (x+h)^(1/3) b = x^(1/3) So on the numerator, now you have (x+h)^(1/3)^3 - x^(1/3)^3 = x+h - x = h And on the denominator you have h( ((x+h)^(1/3))^2 + ((x+h)^(1/3)*x^(1/3)) + (x^(1/3))^2 ) The h's cancel, and you're left with 1/( ((x+h)^(1/3))^2 + ((x+h)^(1/3)*x^(1/3)) + (x^(1/3))^2 ) Plug in 0 for h (since that's where you're taking the limit), and you get your answer (and after doing some algebra to make it all pretty)
Thank you thank you thank you. +rep for you! (as if you need it) I actually tried doing that but I must have made an arithmetic error somewhere because I wasn't getting that answer at all, but it was at like midnight Thanks again! Edit: It appears I have +rep you too much in the past, but I won't forget.