Find the derivative of each function:
(a) $f( x) =2^{x}$ $\quad$ (b) $F( x) =3^{-x}$ $\quad$ (c) $g( x) =\left(\dfrac{1}{2}\right) ^{x^{2}+1}$

Solution (a) $f$ is an exponential function with base $a=2$. $$\begin{eqnarray*} f^\prime (x) =\dfrac{d}{dx}2^{x}= 2^{x}\ln 2\qquad {\color{#0066A7}{\tfrac{d}{dx}a^{x}=a^{x}\ln a}} \end{eqnarray*}$$

(b) Since $F(x)=3^{-x}=\dfrac{1}{3^x}=\left( \dfrac{1}{3}\right) ^{x}$, $F$ is an exponential function with base $\dfrac{1}{3}.$ So, $$\begin{eqnarray*} && F^\prime (x) =\dfrac{d}{dx}\left( \dfrac{1}{3}\right) ^{x} \underset{\underset{{\color{#0066A7}{\tfrac{d}{dx}a^{x}=a^{x}\ln a}}}{\color{#0066A7}{{\uparrow}}}}{=} \left( \dfrac{1}{3}\right) ^{x}\ln \dfrac{1}{3}=\left( \dfrac{1}{3}\right) ^{x}\ln 3^{-1}=-\left( \dfrac{1}{3}\right) ^{x}\ln 3 =-\dfrac{1}{3^{x}}\ln 3 \\ \end{eqnarray*}$$

(c) $y=g( x) =\left( \dfrac{1}{2}\right) ^{x^{2}+1}$ is a composite function. If $u=x^{2}+1,$ then $y=\left( \dfrac{1}{2}\right) ^{u}$ and $$\begin{eqnarray*} && \dfrac{dy}{du} \underset{\underset{{\color{#0066A7}{\tfrac{d}{du}a^{u}=a^{u}\ln a}}}{{\color{#0066A7}{\uparrow}}}}{=} \left(\dfrac{1}{2}\right)^{u}\ln \left( \dfrac{1}{2}\right) \underset{\underset{{\color{#0066A7}{\ln \left( \tfrac{1}{2}\right) =-\ln 2}}}{{\color{#0066A7}{\uparrow}}}}{=} -\left( \dfrac{1}{2}\right) ^{u}\ln 2 \underset{\underset{{\color{#0066A7}{u=x^{2}+1}}}{{\color{#0066A7}{\uparrow}}}}{=} -\left( \dfrac{1}{2}\right) ^{x^{2}+1}\ln 2 \quad \hbox{and}\quad \dfrac{du}{dx}=2x\\ \end{eqnarray*}$$ So, by the Chain Rule, $$\begin{eqnarray*} g^\prime (x) &=&\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot \dfrac{du}{dx}=\left[ -\!\left( \dfrac{1}{2}\right) ^{x^{2}+1}\ln 2\right] (2x) =(-\ln 2)\,{x}\,\!\left( \dfrac{1}{2}\right) ^{x^{2}} \end{eqnarray*}$$