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.
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.
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.
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.
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 

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 

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 

video by PatrickJMT 

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These practice problems require the chain rule.
Basic 

\( f(x) = (x^21)e^{x} \)
Problem Statement 

Calculate the derivative of \( f(x) = (x^21)e^{x} \)
Solution 

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 

video by Krista King Math 

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

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

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 

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 

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 

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 

video by PatrickJMT 

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

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

video by Krista King Math 

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You CAN Ace Calculus
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. 
external links you may find helpful 

WikiBooks  Derivatives of Exponential and Logarithm Functions 
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  doubleangle formulas  

\( \sin(2t) = 2\sin(t)\cos(t)\) 
\(\displaystyle{ \cos(2t) = \cos^2(t)  \sin^2(t) }\) 
Set 4  halfangle 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
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\) 
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\)  
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) 
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = \frac{1}{1+t^2} }\)  
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
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.