Let $f \left(x\right) = {2}^{x}$. Using the chain rule, determine an expression for the derivative of $\left[f \left(g \left(x\right)\right)\right]$.

Apr 26, 2018

If $g \left(x\right)$ is unspecified, the expression for the derivative of $f \left[g \left(x\right)\right]$ is

$\frac{d}{\mathrm{dx}} f \left[g \left(x\right)\right] = \ln 2 \cdot {2}^{g \left(x\right)} \cdot g ' \left(x\right)$.

Explanation:

Using the generic function $g \left(x\right)$, we get

$f \left[g \left(x\right)\right] = {2}^{g \left(x\right)}$

Thus,

$\frac{d}{\mathrm{dx}} f \left[g \left(x\right)\right] = \frac{d}{\mathrm{dx}} {2}^{g \left(x\right)}$

The chain rule says: If $y$ is a function of $u$, and $u$ is a function of $x$, then $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$.

Basically, you treat $g \left(x\right)$ as your variable when differentiating $f$, then use $x$ when differentiating $g$.

$\frac{d}{\mathrm{dx}} f \left[g \left(x\right)\right] = \frac{\mathrm{df}}{\mathrm{dg}} \cdot \frac{\mathrm{dg}}{\mathrm{dx}}$

$\textcolor{w h i t e}{\frac{d}{\mathrm{dx}} f \left[g \left(x\right)\right]} = \frac{d}{\mathrm{dg}} {2}^{g} \cdot \frac{d}{\mathrm{dx}} g \left(x\right)$

$\textcolor{w h i t e}{\frac{d}{\mathrm{dx}} f \left[g \left(x\right)\right]} = \left(\ln 2\right) \left({2}^{g}\right) \cdot g ' \left(x\right)$

$\textcolor{w h i t e}{\frac{d}{\mathrm{dx}} f \left[g \left(x\right)\right]} = \left(\ln 2\right) \left({2}^{g \left(x\right)}\right) g ' \left(x\right)$