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Section3.1Exponential Functions

Motivating Questions
  • What is an exponential function? What important properties does the graph have?

  • How do we take the derivative of an exponential function, and more complicated functions involving sums, products, quotients, and composition with exponential functions?

In this chapter we define exponential functions and their derivatives.

SubsectionExponential Functions

We will start with a formal definition of the exponential function.

Exponential Function
\begin{equation*} f(x)= a \cdot e^x \text{, where } e \approx 2.718281828459... \text{ (an irrational number) and } a \ne 0. \end{equation*}

The constant \(a \) is the \(y\)-value of the \(y\)-intercept of the function.

To understand the exponential function it is good to examine the graph of \(f(x)=e^x\text{.}\)

Figure3.1Graph of \(y = g(x)\text{.}\)

Some important things to note about the function \(f(x)=e^x\text{:}\)

\begin{equation*} f(0)=1, \end{equation*}
\begin{equation*} \lim_{x \to -\infty} e^x=0 \end{equation*}
\begin{equation*} \lim_{x \to \infty} e^x=\infty \end{equation*}

The domain of \(f(x)=e^x\) is \((-\infty,\infty)\) and the range is \((0,\infty)\text{.}\)

SubsectionDerivative of Exponential Functions

Exponential Function

Given an exponential function \(f(x)=ae^x\text{,}\) the derivative is

\begin{equation*} f'(x)= ae^x \end{equation*}

That is \(f(x)=ae^x\) is its own derivative!


Consider the functions \(h(x)=3x^2+5e^x\) and \(g(x)=x^3e^x\text{,}\) find the derivatives of each function.

To find the derivative of \(h(x)\text{,}\) we will use the power rule and the sum or difference rule:

\begin{equation*} h'(x)=6x+5e^x. \end{equation*}

To find the derivative of \(g(x)\text{,}\) we need to use the product rule:

\begin{equation*} g'(x)=\frac{d}{dx}\left[x^3\right]e^x+x^3\frac{d}{dx}\left[e^x\right]=3x^2e^x+x^3e^x. \end{equation*}

Note that the \(e^x\) term appears in both terms since it is its own derivative, thus we can simplify by factoring out the common terms:

\begin{equation*} g'(x)=x^2e^x(3+x). \end{equation*}

Find the derivative of the function \(f(x)=e^{x^3}\text{.}\)

To find the derivative of \(f(x)\) will use the chain rule where the outside function is \(e^x\) and the inside function is \(x^3\text{.}\) Since \(e^x\) is its own derivative we have:

\begin{equation*} f'(x)=e^{x^3}3x^2. \end{equation*}
Exponential Function with Chain Rule

Given an exponential function \(f(x)=ae^{g(x)}\text{,}\) the derivative is

\begin{equation*} f'(x)= ae^{g(x)}g'(x) \end{equation*}

Since \(f(x)=ae^x\) is its own derivative, the original term remains and we multiply by the derivative of the exponent.


Differentiate each of the following functions. State the rule(s) you use, label relevant derivatives appropriately, and be sure to clearly identify your overall answer.

  1. \(p(x) = 5e^{6x}\)

  2. \(\displaystyle m(x) = \frac{e^{3x}}{4x+e^{2x}}\)

  3. \(\displaystyle h(x) = (x^3+5x)e^{-x^2}\)

  4. \(c(x) = \sqrt{5e^{3x}+x}\)

  1. \(p'(x) = 30e^{6x}\text{.}\)

  2. \(m'(x) = \frac{3e^{3x}(4x+e^{2x})-e^{3x}(4+2e^{2x})}{(4x+e^{2x})^2}\text{.}\)

  3. \(h'(x) =(3x^2+5)e^{-x^2}+(x^3+5x)e^{-x^2}(-2x)\text{.}\)

  4. \(c'(x) = \frac{1}{2}(5e^{3x}+x)^{-1/2}(15e^{3x}+1)\text{.}\)

  1. Using the chain rule with an inside function of \(6x\) and an outside function of \(5e^x\) the derivative is

    \begin{equation*} \displaystyle p'(x) = 5e^{6x}(6)=30e^{6x}\text{.} \end{equation*}
  2. Observe that by the quotient rule,

    \begin{equation*} \displaystyle m'(x) = \frac{\frac{d}{dx}\left[e^{3x}\right](4x+e^{2x})-e^{3x}\frac{d}{dx}\left[(4x+e^{2x})\right]}{(4x+e^{2x})^2}\text{.} \end{equation*}

    Applying the chain rule to differentiate \(e^{3x}\) and \(e^{2x}\) we get

    \begin{equation*} m'(x) = \frac{3e^{3x}(4x+e^{2x})-e^{3x}(4+2e^{2x})}{(4x+e^{2x})^2}\text{.} \end{equation*}
  3. By the product rule,

    \begin{equation*} h'(x) = \frac{d}{dx}\left[(x^3+5x)\right]e^{-x^2} + (x^3+5x)\frac{d}{dx}\left[e^{-x^2}\right]\text{.} \end{equation*}

    Applying the chain rule to differentiate \(e^{-x^2}\) we get

    \begin{equation*} h'(x) =(3x^2+5)e^{-x^2}+(x^3+5x)e^{-x^2}(-2x). \end{equation*}
  4. By the chain rule rule

    \begin{equation*} h'(x)=\frac{1}{2}(5e^{3x}+x)^{-1/2}(15e^{3x}+1) \end{equation*}

    Since the derivative of \(5e^{3x}\) is \(5e^{3x}(3)=15e^{3x}\text{.}\)


Find the equation of the line tangent to \(f(x)=4e^{x^2}\) at the point \(x=1\text{.}\)

First find the derivative of \(f(x)\) using the chain rule:

\begin{equation*} f'(x)=4e^{x^2}2x=8xe^{x^2}. \end{equation*}

Then the slope of the tangent is:

\begin{equation*} m=f'(1)=8(1)e^{(1)^2}=8e. \end{equation*}

The y-value the for the tangent line is \(f(1)=4e^{(1^2)}=4e\text{,}\) thus the tangent line is given by:

\begin{equation*} y=8e(x-1)+4e \end{equation*}


  • An exponential function has the form \(f(x) = ae^x\text{.}\) The exponential function \(f(x) = e^x\) has some important properties: \(f(0) = 1 \text{,}\) \(\displaystyle \lim_{x \to -\infty} e^x = 0 \text{,}\) and \(\displaystyle \lim_{x \to \infty} e^x = \infty \text{.}\)

  • The derivative of \(f(x) = ae^x\) is \(f'(x) = ae^x\text{.}\) Combining this with the chain rule, we also see that the derivative of \(f(x) = ae^{g(x)} \) is \(f'(x) = ae^{g(x)}g'(x) \text{.}\)