1. ## Another derivative problem

I have problems with finding derivatives using the definition of a derivative when there are square roots involved for some reason. I'm trying to find the derivative of 1/sqrt b and keep getting the same wrong answer, could someone set me straight about how to approach such a problem?

2. You would look at it like:

$f(b) = b^{-1/2}$

As a square root has an exponent of (1/2), and it being in the denominator makes it negative. From there, it's simple:

$f'(b) = -\frac{1}{2}b^{-3/2}$

or

$f'(b) = -\frac{1}{2\sqrt{b^3}}$

If you mean the limit definition, then:

$f'(b) = \frac{f(b+\Delta b)-f(\Delta b)}{\Delta b}$

as delta-b approaches 0.

&

$f(b) = b^{-1/2}$

so:

$f'(b) = \frac{(b+\Delta b)^{-1/2}-(\Delta b)^{-1/2}}{\Delta b}$

And then the (-1/2) exponents kind of lose me. Sorry. I feel dumb but I don't remember doing this.

3. thanks that actually cleared it up quite a bit for me.

4. Alright then!

I have no clue what to do there really, or at least I'm really tired.