5.15 Derivatives of Exponential Functions

Let

$$\begin{array}{}\text{(a)}& y=a^x,a>0.\end{array}$$

Taking the natural logarithm (= logarithm to the base $e$) of both sides, we get
$$\ln y=x\ln a$$
or
$$\begin{array}{rl}x& =\frac{\ln y}{\ln a}\\ & =\frac{1}{\ln a}\ln y.\end{array}$$
Differentiating with respect to $y$, we get
$$\begin{array}{}\text{(b)}& \frac{dx}{dy}=\frac{1}{\ln a}\cdot \frac{1}{y}.\end{array}$$
It follows from the Derivative Rule for Inverses that
$$\frac{dy}{dx}=\ln a\cdot y$$ or $$\frac{dy}{dx}=\ln a\cdot a^x.$$ Hence, $$\begin{array}{}\text{(c)}& \frac{d}{dx}{a}^x=\ln a\cdot a^x.\end{array}$$ When $a=e$, $\ln a=\ln e=1$, and the formula becomes $$\begin{array}{}\text{(d)}& \frac{d}{dx}{e}^x=e^x.\end{array}$$

Now consider $y=u^v$, where $u$ and $v$ are functions of $x$. To find $y^{\prime }$, we take natural logarithm of both sides:

$$\ln y=v\ln u$$

Now we can use logarithmic differentiation or write the above equation as

$$y=e^{v\ln u},$$

and then apply (d) and the chain rule:

$$\begin{array}{rl}\frac{dy}{dx}& =e^{v\ln u}\frac{d}{dx}(v\ln u)\\ & =u^v[\frac{dv}{dx}\cdot \ln u+v\frac{d}{dx}\ln u]\\ & =u^v[\frac{dv}{dx}\ln u+v(\frac{d}{du}\ln u\cdot \frac{du}{dx})]\\ & =u^v[\frac{dv}{dx}\ln u+v(\frac{1}{u}\cdot \frac{du}{dx})]\\ & =\ln u\cdot u^v\frac{dv}{dx}+v\cdot u^{v-1}\frac{du}{dx}.\end{array}$$

Hence, $$\begin{array}{}\text{(e)}& \frac{d}{dx}{u}^v=\ln u\cdot u^v\frac{dv}{dx}+v\cdot u^{v-1}\frac{du}{dx}\end{array}$$

The derivative of a function with a variable exponent is equal to the sum of the two results obtained by first differentiating, regarding the exponent as constant, and again differentiating regarding the function as constant.