\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \) \( \newcommand{\cm}{\mathrm{cm} } \) \( \newcommand{\sec}{ \, \mathrm{sec} \, } \) \( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \) \( \newcommand{\arccot}{ \, \mathrm{arccot} \, } \) \( \newcommand{\arcsec}{ \, \mathrm{arcsec} \, } \) \( \newcommand{\arccsc}{ \, \mathrm{arccsc} \, } \) \( \newcommand{\sech}{ \, \mathrm{sech} \, } \) \( \newcommand{\csch}{ \, \mathrm{csch} \, } \) \( \newcommand{\arcsinh}{ \, \mathrm{arcsinh} \, } \) \( \newcommand{\arccosh}{ \, \mathrm{arccosh} \, } \) \( \newcommand{\arctanh}{ \, \mathrm{arctanh} \, } \) \( \newcommand{\arccoth}{ \, \mathrm{arccoth} \, } \) \( \newcommand{\arcsech}{ \, \mathrm{arcsech} \, } \) \( \newcommand{\arccsch}{ \, \mathrm{arccsch} \, } \)

17Calculus Derivatives - Exponential Functions

Limits

Using Limits


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Improper Integrals

Trig Integrals

Length-Area-Volume

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Practice

Calculus 1 Practice

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This is one of the easiest rules you will learn.

Basic Exponential Rule

\(\displaystyle{ \frac{d}{dt}[e^t] = e^t }\)

Exponential With Chain Rule

\(\displaystyle{ \frac{d}{dt}[e^u] = e^u \frac{du}{dt} }\)

It looks like we didn't do anything here. However, the exponential function is the only function whose derivative is itself.
Before we go on, let's watch a video that gives an intuitive explanation of the derivative of exponential functions and why \(f(x)=e^x\) is special.

3Blue1Brown - Derivatives of exponentials [13min-49secs]

video by 3Blue1Brown

Okay, so what do you do if you have a base other than \(e\)? The formula is fairly straightforward but let's derive from our rules of logarithms. Since we don't like to memorize formulas and we already know the logarithm rules, why not just derive it when we need it, since we don't use it very much?

So, let's convert \(y=a^x\), where \(a\) is a constant, into a form with \(e\).
\( \begin{array}{rcl} y & = & a^x \\ \ln(y) & = & \ln(a^x) \\ & = & x\ln(a) \\ e^{\ln(y)} & = & e^{x\ln(a)} \\ y & = & e^{x\ln(a)} \end{array} \)

So now when we take the derivative of \(y = a^x\), we can actually take the derivative of \(y=e^{x\ln(a)}\). Using the chain rule, we have \((a^x)' = (\ln(a))e^{x\ln(a)} \). Notice how didn't have to memorize this formula. We used the logarithm rules we already know.

This next video goes through all the explanation again. It is always good to get explanations from different sources since it will help you understand the material better.

PatrickJMT - Derivatives of Exponential Functions [5min-49secs]

video by PatrickJMT

Here is an interesting video that shows how to get the equation for the derivative of \(f(x)=a^x\) another way. He shows that \(\displaystyle{f'(x)=\frac{d[a^x]}{dx}=a^x f'(0)}\). This is an interesting and unusual way to think about the derivative.

Dr Chris Tisdell - Derivative of exponentials [9min-15secs]

video by Dr Chris Tisdell

So far, we've only been looking at equations with exponential functions. Here is a video discussing the graph, the derivative and the tangent line of three exponential functions. This helps you get more of an intuitive feel for this function and it's derivative.

MathTV - Some Natural Exponential Functions and Tangent Lines [4min-11secs]

video by MathTV

Practice

Unless otherwise instructed, calculate the derivative of these functions.

Here are a few practice problems that do not require the chain rule.

\( y = e^x(x+x\sqrt{x}) \)

Problem Statement

Calculate the derivative of \( y = e^x(x+x\sqrt{x}) \)

Solution

1069 video

video by Krista King Math

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\( f(x) = 4^x+3e^x+x^4 \)

Problem Statement

Calculate the derivative of \( f(x) = 4^x+3e^x+x^4 \)

Solution

1074 video

video by PatrickJMT

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\( f(x) = e^x x^2 \)

Problem Statement

Calculate the derivative of \( f(x) = e^x x^2 \)

Solution

1076 video

video by PatrickJMT

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These practice problems require the chain rule.

Basic

\( f(x) = (x^2-1)e^{-x} \)

Problem Statement

Calculate the derivative of \( f(x) = (x^2-1)e^{-x} \)

Solution

1066 video

video by Krista King Math

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\(\displaystyle{ f(x)=xe^{\sqrt{x}} }\)

Problem Statement

Calculate the derivative of \(\displaystyle{ f(x)=xe^{\sqrt{x}} }\)

Solution

1067 video

video by Krista King Math

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\(\displaystyle{ f(x) = \frac{1-e^{-x}}{x} }\)

Problem Statement

Calculate the derivative of \(\displaystyle{ f(x) = \frac{1-e^{-x}}{x} }\)

Solution

1068 video

video by Krista King Math

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\(\displaystyle{ 3e^{ x^2+7 } }\)

Problem Statement

Calculate the derivative of this function and give your final answer in completely factored form. \(\displaystyle{ 3e^{ x^2+7 } }\)

Final Answer

\(\displaystyle{ \frac{d}{dx} \left[ 3e^{ x^2+7 } \right] = 6xe^{x^2+7} }\)

Problem Statement

Calculate the derivative of this function and give your final answer in completely factored form. \(\displaystyle{ 3e^{ x^2+7 } }\)

Solution

\(\displaystyle{ \frac{d}{dx} \left[ 3e^{ x^2+7 } \right] }\)

\(\displaystyle{ 3e^{x^2+7} \cdot \frac{d}{dx}[x^2+7] }\)

\( 3e^{x^2+7} \cdot (2x) \)

\( 6xe^{x^2+7} \)

Another way to work this is with the substitution method.

let \(u=x^2+7\)

\(\displaystyle{ \frac{d}{dx} \left[ 3e^{ x^2+7 } \right] }\)

\(\displaystyle{ 3\frac{d[e^u]}{du} \cdot \frac{d[x^2+7]}{dx} }\)

\( 3e^u \cdot (2x) \)

\( 6xe^{x^2+7} \)

Final Answer

\(\displaystyle{ \frac{d}{dx} \left[ 3e^{ x^2+7 } \right] = 6xe^{x^2+7} }\)

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Intermediate

\(\displaystyle{ f(x) = e^{x\sin(2x)} }\)

Problem Statement

Calculate the derivative of \(\displaystyle{ f(x) = e^{x\sin(2x)} }\)

Solution

1077 video

video by PatrickJMT

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\(\displaystyle{ g(x) = 2e^{\cos(x)\sin(5x)} }\)

Problem Statement

Calculate the derivative of \(\displaystyle{ g(x) = 2e^{\cos(x)\sin(5x)} }\)

Solution

1075 video

video by PatrickJMT

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\(\displaystyle{f(t)=\cos\left(2^{\pi t}\right) }\)

Problem Statement

Calculate the derivative of \(\displaystyle{f(t)=\cos\left(2^{\pi t}\right) }\)

Solution

1078 video

video by PatrickJMT

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For what values of \( x \) does \( h(x)=5e^{5x}-25x \) have negative derivatives?

Problem Statement

For what values of \( x \) does \( h(x)=5e^{5x}-25x \) have negative derivatives?

Solution

1079 video

video by PatrickJMT

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\(\displaystyle{ y = \cos \left( \frac{1-e^{2x}}{1+e^{2x}} \right) }\)

Problem Statement

Calculate the derivative of this function and give your final answer in completely factored form. \(\displaystyle{ y = \cos \left( \frac{1-e^{2x}}{1+e^{2x}} \right) }\)

Solution

986 video

video by Krista King Math

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You CAN Ace Calculus

Topics You Need To Understand For This Page

basic derivative rules

power rule

product rule

quotient rule

For the basic exponential derivatives you do not need the chain rule. But we discuss it on this page. Each section is labeled. So if you have not studied the chain rule yet, you can read the sections that apply to you and then come back here once you have studied it.

Related Topics and Links

Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\)

\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\)

\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\)

\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\)

Set 2 - squared identities

\( \sin^2t + \cos^2t = 1\)

\( 1 + \tan^2t = \sec^2t\)

\( 1 + \cot^2t = \csc^2t\)

Set 3 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle formulas

\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\)

\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\)

Trig Derivatives

\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\)

 

\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\)

\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\)

 

\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\)

\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\)

 

\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\)

Inverse Trig Derivatives

\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\)

 

\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\)

\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\)

 

\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\)

\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

 

\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

Trig Integrals

\(\int{\sin(x)~dx} = -\cos(x)+C\)

 

\(\int{\cos(x)~dx} = \sin(x)+C\)

\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\)

 

\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)

\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\)

 

\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\)

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Practice Instructions

Unless otherwise instructed, calculate the derivative of these functions.

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