f'(x)... the hard way (limit questions)

[Benjamin753]

Hi anybody...

 

I have a problem. Someone (or most) should find it really easy, but I'm having trouble. Here it is.

 

f(x)= x^(1/3) (or the cube root of x)

need to find f'(x) using limit as h->0 of (f(x+h) - f(x))/h

 

Now, I already know what the answer is: (1/(3(x^(2/3))), but how do I get that using the difference quotient (above).

 

Obviously, my algebra skills are rusty, but here is what I did to get the wrong answer:

 

((x+h)^(1/3)) - (x^(1/3)) ( times the conjugate)

------------------------

h

 

 

((x+h)^(2/3)) + (x^(2/3))

-------------------------

((x+h)^(2/3)) + (x^(2/3)) =

 

 

x + h - x

---------------------------

h((x+h)^(2/3)) + (x^(2/3)) =

 

 

1

----------------------------

((x+h)^(2/3)) + (x^(2/3))

 

(then sub 0 for h, because it is a limit problem) then ---->

 

f'(x) =

 

1

------------

2(x^(2/3))

 

which is not correct...

 

Anybody help me out with this one?? I would appreciate it!

 

Thanks!

[Benjamin753]

Nevermind... I got it... my conjugate was wrong... oops.

[Dave]

FYI, there was a complete thread on this here.