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

Mar 27, 2017

$\frac{1}{2 \setminus \sqrt{{e}^{2 x} - 2}}$

#### Explanation:

$f ' \left(g \left(x\right)\right)$ means that we find $f ' \left(x\right)$ and substitute $g \left(x\right)$ for $x$.

Now, in order to solve $f ' \left(x\right)$, we need to use the chain rule since it is a composite function. We recall the chain rule: $\frac{\mathrm{du}}{\mathrm{dx}} = \frac{\mathrm{du}}{\mathrm{dv}} \cdot \frac{\mathrm{dv}}{\mathrm{dx}}$, where $u$ and $v$ are functions of $x$. For this function, $\setminus \sqrt{x - 2}$, we say that $u$ is $\setminus \sqrt{x - 2}$ and $v$ is $x - 2$.

In order to find $\frac{\mathrm{du}}{\mathrm{dx}}$, we first find $\frac{\mathrm{du}}{\mathrm{dv}}$, which is $\frac{d \left(\setminus \sqrt{x - 2}\right)}{d \left(x - 2\right)} = \frac{1}{2} \cdot {\left(x - 2\right)}^{- \frac{1}{2}} = \frac{1}{2 \setminus \sqrt{x - 2}}$. Next, we find $\frac{\mathrm{dv}}{\mathrm{dx}}$, which is $\frac{d \left(x - 2\right)}{\mathrm{dx}} = 1$. Multiplying these gives us $f ' \left(x\right)$, $\frac{1}{2 \setminus \sqrt{x - 2}}$.

The only thing now to do is to substitute $g \left(x\right)$ in for $x$. We get the final answer: $\frac{1}{2 \setminus \sqrt{{e}^{2 x} - 2}}$.