# Homework Help: Simple Single Variable Derivative Question

1. Oct 28, 2005

### cappygal

wow! I am having a major brain fart here .. I'm in multivariable calc but I need to take the derivative of a single variable function using the definition of derivative. The function is:

f(x)=abs(x)^(3/2) and I need to find the derivative when a=0

so by the def of derivative, the derivative equals:

lim (h->0) of [abs(a+h)^(3/2)-abs(a)^3/2]/h

I know I'm not allowed to just "plug in" a=0 to get the derivative at this point .. but how do I do it? I can't combine the absolute value terms because of the triangle inequality (for ex. abs(-6)-abs(3)=3 but abs(-6-3)=9)

help! I feel really stupid right now. :grumpy:

2. Oct 28, 2005

### quasar987

why can't you plug a=0 ?:grumpy:

3. Oct 28, 2005

### cappygal

I thought when you were using definition of derivitive, you aren't allowed to plug in for a until you have it simplified? Am I wrong? I could be .. it's been a year since I've had to work with definition of derivative.

4. Oct 28, 2005

### hypermorphism

There's no such rule. "a" is a placeholder for a constant; place your constant in there to get the derivative there. Leave it as "a" to get an expression for the derivative (if it exists) at all values of "a".

5. Oct 28, 2005

### benorin

use the *other* definition of limit:
f '(a) = lim (x->a) of [abs(x)^(3/2)-abs(a)^3/2]/(x-a)

then factor the top as a difference of cubes.

6. Oct 28, 2005

### HallsofIvy

You cannot plug in h= 0 because that would give 0/0. But if you are finding the derivative at a given x= a, you certainly can plug in that value of a before you start. You are looking for
$$lim_{h\rightarrow0}\frac{h^{\frac{3}{2}}}{h}$$
That should be easy.