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Section 6.2 Exponential Functions & Derivatives
Objectives
Identify the general exponential function \(y=b^x\) for any base \(b >0,b\neq 1\) and its basic properties (domain, range, graphical behavior depending on value of \(b\) ), and apply rules of exponents to exponential functions
Evaluate limits of functions involving exponential expressions and analyze the end behavior of those functions
Sketch and identify transformations of exponential functions, especially as examples of exponential growth/decay in real-world contexts
Recognize and interpret the number \(e\) in terms of limits, derivatives, and geometry
Compute derivatives and antiderivatives of functions involving natural exponential expressions and use qualitative techniques to analyze key behavior of those functions (extrema, concavity, etc.)
Subsection 6.2.1 Before Class
https://mymedia.ou.edu/media/6.2-1/1_gj4wt9ztFigure 51. Pre-Class Video 1
Subsubsection 6.2.1.1 Exponential Functions
Definition 6.2.1 . Exponential Function.
An exponential function is a function of the form \(f(x) = b^x\text{,}\) where \(b\) is a positive constant.
Properties of Exponential Functions. Let \(f(x) = b^x\)
Domain: \((-\infty,\infty)\)
Range: \([0,\infty)\)
If \(b\gt 1\text{,}\) then \(f(x)\) is increasing
If \(0\lt b \lt 1\text{,}\) then \(f(x)\) is decreasing
\(\displaystyle \ds \lim_{x\to \infty} f(x) =
\begin{cases}
\infty & \text{if } b\gt 1 \\
0 & \text{if } 0\lt b\lt 1
\end{cases}\)
\(\displaystyle \ds \lim_{x\to -\infty} f(x) =
\begin{cases}
0 & \text{if } b\gt 1 \\
\infty & \text{if } 0\lt b\lt 1
\end{cases}\)
Example 6.2.2 .
For the following functions, find the limits and sketch the graph.
\(\displaystyle f(x) = 2(1.2^x)+3\)
\(\displaystyle g(x) = 3^{-x} -1\)
Solution .
\(\ds \lim_{x\to -\infty} f(x) = 3\) and \(\ds \lim_{x\to\infty} f(x) = \infty\)
\(\ds \lim_{x\to -\infty} g(x) = \infty\) and \(\ds \lim_{x\to\infty} g(x) = -1\)
Definition 6.2.3 . Euler’s Constant (\(e\) ).
\(e\) is defined to the number for which \(\ds \lim_{h\to 0} \dfrac{e^h-1}{h}=1\)
Subsubsection 6.2.1.2 Calculus of Exponentials
Derivative of an Exponential (First Attempt). If \(f(x) = b^x\text{,}\) then \(f'(x) = f'(0)b^x\)
Proof.
Using the definition of the derivative, we have
\begin{equation*}
\ds f'(x) = \lim_{h\to 0} \dfrac{b^{x+h}-b^x}{h} = \lim_{h\to 0} \dfrac{b^x(b^h-1)}{h} = b^x\cdot \lim_{h\to 0} \dfrac{b^h-1}{h}
\end{equation*}
Now,
\begin{equation*}
\ds f'(0) = \lim_{h\to 0} \dfrac{f(0+h)-f(0)}{h} = \lim_{h\to 0} \dfrac{b^{0+h}-b^0}{h} = \lim_{h\to 0} \dfrac{b^h-1}{h}
\end{equation*}
So we conclude that \(f'(x) = f'(0)b^x\)
This means we have the following interpretation of \(f(x) = e^x\text{:}\)
Special Meaning of \(e\) .
\(f(x) = e^x\) is the unique exponential function whose tangent line at the point \((0,1)\) is exactly 1, i.e. \(f'(0) = 1\)
Derivative of \(e^x\) . \(\dfrac{d}{dx}\left[e^x\right] = e^x\)
Antiderivative of \(e^x\) . \(\ds \int e^x\, dx = e^x + C\)
Subsection 6.2.2 Pre-Class Activities
Example 6.2.4 .
Write the domain of the function:
\(\displaystyle f(x) = \dfrac{1-e^{x^2}}{1-e^{4-x^2}}\)
\(\displaystyle g(x) = \dfrac{1+x}{3^{\sin x}}\)
\(\displaystyle h(t) = \sqrt{4^t - 16}\)
Solution .
\(\displaystyle (-\infty,-2)\cup (-2,2)\cup (2,\infty)\)
\(\displaystyle (-\infty,\infty)\)
\(\displaystyle [2,\infty)\)
Example 6.2.5 .
Find the indicated limit:
\(\displaystyle \ds \lim_{x\to \infty} (1.0001)^x\)
\(\displaystyle \ds \lim_{x\to \infty} \dfrac{e^{3x}-e^{-3x}}{e^{3x} + e^{-3x}}\)
\(\displaystyle \ds \lim_{x\to \infty} (e^{-2x}\sin x)\)
Solution .
\(\displaystyle \ds \lim_{x\to \infty} (1.0001)^x = \infty\)
\(\displaystyle \ds \lim_{x\to \infty} \dfrac{e^{3x}-e^{-3x}}{e^{3x} + e^{-3x}} = 1\)
\(\displaystyle \ds \lim_{x\to \infty} (e^{-2x}\sin x) = 0\)
Example 6.2.6 .
Find the derivative of the function:
\(\displaystyle f(x) = e^4\)
\(\displaystyle g(r) = e^r + r^e\)
\(\displaystyle f(x) = \dfrac{e^x}{1+e^x}\)
Solution .
\(\displaystyle f'(x) = 0\)
\(\displaystyle g'(r) = e^r + er^{e-1}\)
\(\displaystyle f'(x) = \dfrac{(1+e^x)(e^x)-(e^x)(e^x)}{(1+e^x)^2}\)
Example 6.2.7 .
Find the equation of the tangent line to the curve \(y = xe^x\) at the point \((1,e)\text{.}\)
Question 6.2.8 .
Use this space to write any questions or concerns you have from the pre-class portion of this section.
Subsection 6.2.3 In Class
Subsubsection 6.2.3.1 Examples
Example 6.2.9 .
Compute \(f'(x)\text{,}\) if \(f(x) = e^{\tan x}\)
Solution . \(f'(x) = e^{\tan x}\cdot \sec^2 x\)
Example 6.2.10 .
Compute \(f'(x)\text{,}\) if \(f(x) = \tan(e^x)\)
Example 6.2.11 .
Find \(y'\) if \(y=e^{-6x}\cos(2x)\)
Solution . \(y'=(-6e^{-6x})(\cos 2x) +(e^{-6x})(-2\sin 2x)\)
Example 6.2.12 .
Find the absolute maximum and absolute minimum of \(y=xe^{-x}\)
Solution . Absolute max is at \(x=1\) and there is no absolute min
Example 6.2.13 .
Find \(\dfrac{dy}{dx}\text{,}\) if \(e^{x/y} = y-x\)
Solution . \(\dfrac{dy}{dx} = \dfrac{ye^{x/y}+y^2}{y^2+xe^{x/y}}\)
Example 6.2.14 .
Compute the derivatives:
\(\displaystyle y = x^2e^{-1/x}\)
\(\displaystyle g(x) = e^{x^2-x}\)
\(\displaystyle f(t) = \sqrt{1+te^{-2t}}\)
Solution .
\(\displaystyle y'= 2xe^{-1/x} + (x^2)\lrpar{\dfrac{1}{x^2}e^{-1/x}}\)
\(\displaystyle g'(x) = (2x-1)e^{x^2-x}\)
\(\displaystyle f'(t) = \dfrac{1}{2}(1+te^{-2t})^{-1/2}(e^{-2t}-2te^{-2t})\)
Example 6.2.15 .
Find the absolute maximum and absolute minimum of \(f(x) = xe^{-x^2/8}\) on \([-1,4]\)
Solution . Absolute max occurs at \(\lrpar{2,2e^{-1/2}}\) and aboslute min occurs at \(\lrpar{-1,-e^{-1/8}}\)
Example 6.2.16 .
Evaluate the integral:
\(\displaystyle \ds \int_0^1 (x^e + e^x)\, dx\)
\(\displaystyle \ds\int x^3e^{x^4}\, dx\)
\(\displaystyle \ds\int e^x\sqrt{1+e^x}\, dx\)
Solution .
\(\displaystyle \ds \int_0^1 (x^e + e^x)\, dx = \dfrac{1}{e+1}+e-1\)
\(\displaystyle \ds\int x^3e^{x^4}\, dx = \dfrac{1}{4}e^{4x}+C\)
\(\displaystyle \ds\int e^x\sqrt{1+e^x}\, dx = \dfrac{2}{3}(1+e^x)^{3/2} + C\)
Example 6.2.17 .
Compute \(\ds \int_1^2 \dfrac{e^{1/x}}{x^2}\, dx\)
Solution . \(\ds \int_1^2 \dfrac{e^{1/x}}{x^2}\, dx = e-e^{1/2}\)
Example 6.2.18 .
Find \(f(x)\) if \(f''(x) = 3e^x+5\sin x\text{,}\) \(f(0)=1\text{,}\) and \(f'(0)=2\)
Example 6.2.19 .
The error function , \(\ds \text{erf}(x) = \dfrac{2}{\sqrt{\pi}}\int_0^x e^{-t^2}\,dt\) is a useful function in probability, statistics, and engineering. Show that
\begin{equation*}
\ds \int_a^b e^{-t^2}\, dt = \dfrac{1}{2}\sqrt{\pi}\left[\text{erf}(b) - \text{erf}(a)\right]\text{.}
\end{equation*}
Solution .
Rearranging, we have
\begin{equation*}
\dfrac{\sqrt{\pi}}{2}\text{erf}(x) = \ds \int_0^x e^{-t^2}\, dt
\end{equation*}
So, \(\ds \int_0^a e^{-t^2}\, dt = \dfrac{\sqrt{\pi}}{2}\text{erf}(a)\) and \(\ds \int_0^b e^{-t^2}\, dt = \dfrac{\sqrt{\pi}}{2}\text{erf}(b)\text{.}\) Now, using properties of integrals, we have
\begin{equation*}
\ds \int_a^b e^{-t^2}\, dt = -\int_0^a e^{-t^2}\, dt + \int_0^b e^{-t^2}\,dt = \dfrac{1}{2}\sqrt{\pi}\left[\text{erf}(b) - \text{erf}(a)\right]
\end{equation*}
Subsection 6.2.4 After Class Activities
Example 6.2.20 .
Show that the function \(y = e^x + e^{-x/2}\) satisfies the differential equation \(2y'' - y' - y = 0\)
Solution .
Compute \(y'\) and \(y''\text{:}\)
\begin{equation*}
y' = e^x -\dfrac{1}{2}e^{-x/2}
\end{equation*}
\begin{equation*}
y'' = e^x + \dfrac{1}{4}e^{-x/2}
\end{equation*}
Now, plugging in for \(y''\) and \(y'\text{,}\) we have
\begin{equation*}
2(e^x + \dfrac{1}{4}e^{-x/2}) - (e^x -\dfrac{1}{2}e^{-x/2}) - (e^x + e^{-x/2})
\end{equation*}
which does equal 0.
Example 6.2.21 .
Find an equation of the tangent line to the curve \(xe^y + ye^x = 1\) at the point \((0,1)\)
Example 6.2.22 .
Compute \(\dfrac{d^{1000}}{dx^{1000}}\left[xe^{-x}\right]\)
Solution . \(\dfrac{d^{1000}}{dx^{1000}}\left[xe^{-x}\right] = -1000e^{-x} +xe^{-x}\)
Example 6.2.23 .
If \(f(x) = 3 + x + e^x\text{,}\) find \((\inv{f})'(4)\)
Example 6.2.24 .
Evaluate \(\ds \lim_{x\to \pi} \dfrac{e^{\sin x} - 1}{x-\pi}\)
Solution . \(\ds \lim_{x\to \pi} \dfrac{e^{\sin x} - 1}{x-\pi} = -1\)
Example 6.2.25 .
Find the volume of the solid obtained by rotating the region bounded by the curves \(y = e^x\text{,}\) \(y = 0\text{,}\) \(x = 0\text{,}\) and \(x = 1\) about the \(x-\) axis.
Solution . The volume is \(\dfrac{\pi}{2}[e^2-1]\)
