If f(x) =-sqrt(2x-1)  and g(x) = (2-1/x)^2 , what is f'(g(x)) ?

Nov 29, 2016

The question is asking, "what is the equation for the derivative of $f$, when its input is $g \left(x\right)$?

So we need to do two things: (1) Find the general derivative of $f$, and (2) Plug in $g \left(x\right)$ as the input to this derivative.

Step (1):
If
$f \left(x\right) = - \sqrt{2 x - 1} = - {\left(2 x - 1\right)}^{1 / 2}$,
then
$f ' \left(x\right) = - \frac{1}{2} {\left(2 x - 1\right)}^{- 1 / 2} \left(2\right)$
$\implies f ' \left(x\right) = - \frac{1}{\sqrt{2 x - 1}}$
by use of both the power rule and the chain rule.

This is the general derivative of $f$, with respect to some input $x$. If we plug in different inputs, we get different outputs.

Step (2):
Now, we plug in the input $g \left(x\right)$ (instead of just $x$) to the derivative $f ' :$
$f ' \left(\textcolor{red}{x}\right) = - \frac{1}{\sqrt{2 \textcolor{red}{x} - 1}}$

$\implies f ' \left[\textcolor{b l u e}{g} \left(x\right)\right] = - \frac{1}{\sqrt{2 \textcolor{b l u e}{{\left(2 - \frac{1}{x}\right)}^{2}} - 1}}$

$\implies f ' \left[g \left(x\right)\right] = - \frac{1}{\sqrt{2 {\left(2 - \frac{1}{x}\right)}^{2} - 1}}$.

Depending on your teacher, this is likely as far as you'll need to go.

Bonus:

Please note: The domain of $f ' \left(x\right) = - \frac{1}{\sqrt{2 x - 1}}$ is $\left\{x | 2 x - 1 > 0\right\}$, which is $x > \frac{1}{2}$.
(Slightly more limiting than the domain of just $f \left(x\right)$, which is $x \ge \frac{1}{2}$)

The domain of $g \left(x\right) = {\left(2 - \frac{1}{x}\right)}^{2}$ is $\left\{x | x \ne 0\right\}$.

That means the domain of this composite function $f ' \left[g \left(x\right)\right]$ has to satisfy both of these restrictions. Luckily, that's pretty easy, considering the domain of $f ' \left(x\right)$ is a subset of the domain of $g \left(x\right)$. So the domain of $f ' \left[g \left(x\right)\right]$ is
$\left\{x | x > \frac{1}{2} \cap x \ne 0\right\}$, or just $\left\{x | x > \frac{1}{2}\right\}$.