# Finding the derivitive using limits?

1. Sep 29, 2008

### ne12o

Hi,
I've stumbled across a question. It must be solved using limits. I just don't know where to start. After doing the f(x+h) - f (x) / h i end up with a cluster that cannot be simplified... any help would be great.

the equation is as follows.

y = cube root( X^4 - 3X)

2. Sep 30, 2008

### Tac-Tics

Very interesting. I'm not sure how to prove this either, but I'm really interested in seeing the solution!

I think the core difficulty in this problem is with the root, so let's simplify the problem. How do we find f' for f(x) = sqrt(x)?

Then, we can probably generalize to n-th roots and use the chain rule to finish it off.

My guess is that you have to abuse the relation $$\sqrt[n]{x} = e^{\ln(x^{\frac{1}{n}})} = e^{n \ln(x)}$$. If you were a brutish proof person, you could probably do something with the taylor series of $$e^x$$ and ln(x) to solve it with limits.

Last edited: Sep 30, 2008
3. Sep 30, 2008

The difference of two cubes can be factored this way:

$$A^3 - B^3 = (A-B)(A^2 + AB + B^2)$$

This works for any quantities $$A, B$$, so (curse the database errors which prevent preview from working)

$$a-b = \left(a^{1/3} - b^{1/3}\right)\left(a^{2/3} + a^{1/3} b^{1/3} + b^{1/3}\right)$$

The numerator of the difference quotient has

\begin{align*} a & = \left((x+h)^4 - 3(x+h)\right)^{1/3}\\ b & = \left(x^4 - 3x \right)^{1/3} \end{align*}

To rationalize the difference quotient multiply by this fraction:

$$\frac{\left((x+h)^4 - 3(x+h)\right)^{2/3} + \left((x+h)^4 - 3(x+h)\right)^{1/3} \left(x^4 - 3x\right)^{1/3} + \left(x^4 - 3x\right)^{2/3}}{\left((x+h)^4 - 3(x+h)\right)^{2/3} + \left((x+h)^4 - 3(x+h)\right)^{1/3} \left(x^4 - 3x\right)^{1/3} + \left(x^4 - 3x\right)^{2/3}}$$

this will leave you with (use the second factoring formula from the top of my post)

$$\frac{\left((x+h)^4 - 3(x+h)) - (x^4 - 3x)\right)}{h\left((x+h)^4 - 3(x+h)\right)^{2/3} + \left((x+h)^4 - 3(x+h)\right)^{1/3} \left(x^4 - 3x\right)^{1/3} + \left(x^4 - 3x\right)^{2/3}}$$

This still isn't pretty, but the work of taking the limit can now be performed.

4. Sep 30, 2008

### ne12o

whoa thanks... now trying to understand it.

thanks!