# Taking the derivative of a compound fractions

So I need to do a bunch of stuff with the derivative of $$f(x) = \frac{1}{\sqrt{x}} .$$

Lack of sleep is murdering my ability to do compound fractions today. Can someone show me how to either rationalize this or take its derivative?

-
What you have is NOT a compound fraction. (I didn't know that term, I checked wikipedia just now.) The function can be written as $\frac{1}{\sqrt{x}} = \frac{1}{x^{1/2}} = x^{-1/2}$. Do you know how to differentiate $x^n$? –  Srivatsan Sep 15 '11 at 22:23
Yup, I messed up terminology. I meant to say complex fraction rather than compound. I knew I was forgetting something really simple. Thanks. –  bucketmouse Sep 16 '11 at 1:07

We know that differentiation of $x^n$ ($n \in \mathbb{R}$) with respect to $x$ is $n \times x^{n-1}$.

Notice that we could write: $$\frac{1}{\sqrt{x}} = \frac{1}{x^{1/2}} = x^{-1/2}$$ now applying differentiation:

$$\frac{\mathrm{d} }{\mathrm{d} x} \bigl(x^{-1/2}\bigr)= -\frac{1}{2} \times x^{-1/2-1} = -\frac{1}{2} \times x^{-3/2}=-\frac{1}{2 \sqrt{x^{\large3}}}$$.

-

More generally, the chain rule is your friend:

$$(f(g(x))' = f'(g(x)) g'(x).$$

It's a bit much in your case, but setting $f(x) = 1/x$ and $g(x) = \sqrt x$, since $f'(x) = -1/x^2$ and $g'(x) = x^{-1/2}/2$, $$(1/\sqrt x)' = (-1/(\sqrt x)^2) (x^{-1/2}/2) = -x^{-3/2}/2.$$

For more complicated expressions, you just have to recursively apply the chain rule until you run out of functions to compose and differentiate. I find it remarkably easy to make errors doing this, so I always go slowly and check each step. As a matter of fact, in my first version of this, I left out the "-" in $(-1/(\sqrt x)^2)$.

-