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?
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?
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 $(-1/(\sqrt x)^2)$.