# Derivative of a square root fraction. HELP!

1. Sep 11, 2009

### curlybit89

1. The problem statement, all variables and given/known data

What is f'(x) of f(x) 1/sqrt(2x)?

2. The attempt at a solution

In applying the problem to the derivative formula:

(1 / sqrt(2(x + h)) - 1 / sqrt(2x)) / h

I multiplied the problem by a special form of one but that only put the rationals on the bottom of the division. This looked too messy.

Alternatively, multiplying each side of the first division by it's denominator yielded the following:

(sqrt(2x) - sqrt(2x2h) / sqrt(2x^2 * 2x * 2h)) / h

I'm not sure how to proceed from here.

Another alternative I attempted was to use the exponent method which for me yielded:

original: 2x^-(1/2) = -(x)^-(3/2)

This answer did not seem appropriate.

2. Sep 11, 2009

### CFDFEAGURU

Think of another way to write the derivate. What is another way to write 1/sqrt(2x) ?

Thanks
Matt

3. Sep 11, 2009

### Jonathan G

Easiest solution in my opinion would be to just re-write it as $$(2x)^{-1/2}$$ and simply take the derivative of that.

Don't forget it is f'(x)=n*$$(u)^{n-1}$$ *u'

4. Sep 11, 2009

### Elucidus

You are almost correct here.

$$\frac{1}{\sqrt{2x}} = (2x)^{-1/2}$$

$$\frac{d}{dx}\left( (2x)^{-1/2} \right) = -\frac{1}{2}(\bold{2}x)^{-3/2}(2)$$

Proceed from there...

--Elucidus

5. Sep 11, 2009

### VietDao29

You are missing a plus, and a pair of parentheses there.

When solving limit of the Indeterminate Form 0/0, you should try to factorize it. And factorizing the numerator, which contains radicals is impossible, right? So, one should think of a way to rationalize it. How about multiplying both numerator and denominator by

$$\sqrt{2x} + \sqrt{2x + 2h}$$?