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The logarithm rule is not as simple as the exponential derivative but it is still very straightforward.
Basic Logarithm Rule 

\(\displaystyle{ \frac{d}{dt}[\ln(t)] = \frac{1}{t} }\) 
Logarithm With Chain Rule 

\(\displaystyle{ \frac{d}{dt}[\ln(u)] = \frac{1}{u}\frac{du}{dt} }\) 
It is probably not clear just from the equation that the derivative of \(\ln(x)\) is \(1/x\). Here is a great video explaining, first intuitively, then from the limit, where this derivative comes from. Although he says you can stop the video after the intuitive explanation, watching the entire video will help you a lot (it's not very long).
video by MathTV 

Many times, it helps to simplify a logarithmic expression before taking the derivative. Here are a few rules that should help you.
\( \ln(xy) = \ln(x) + \ln(y) \) 
\( \ln(x/y) = \ln(x)  \ln(y) \) 
\( \ln(x^y) = y \ln(x) \) 
Here is a short video clip that goes through these equations again.
video by PatrickJMT 

Before jumping into some practice problems, take a couple of minutes to watch this next video. It will help you see some common mistakes that you can avoid when taking the derivative of logarithm functions.
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.
\(\displaystyle{f(x)=\frac{1}{\ln(x)}}\)
Problem Statement 

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

video by Krista King Math 

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\(f(x)=\sqrt{x}\ln(x)\)
Problem Statement 

Calculate the derivative of \(f(x)=\sqrt{x}\ln(x)\)
Solution 

video by PatrickJMT 

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\(\displaystyle{y=\frac{\ln x}{1+\ln(2x)}}\)
Problem Statement 

Calculate the derivative of \(\displaystyle{y=\frac{\ln x}{1+\ln(2x)}}\)
Solution 

Although the chain is not required here, the instructor in this video does use the chain rule to calculate the derivative. You can use a logarithm rule to get \(\ln(2x) = \ln(2)+\ln(x)\) before taking the derivative, in order to avoid the chain rule.
video by PatrickJMT 

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\(y=\ln(x)+2^x+\sin(x)\)
Problem Statement 

Calculate the derivative of \(y=\ln(x)+2^x+\sin(x)\)
Solution 

video by PatrickJMT 

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These practice problems require the chain rule.
Basic 

\(f(x)=\ln(x^2+10)\)
Problem Statement 

Calculate the derivative of \(f(x)=\ln(x^2+10)\)
Solution 

video by PatrickJMT 

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\(y=\ln(x^2+x)\)
Problem Statement 

Calculate the derivative of \(y=\ln(x^2+x)\)
Solution 

video by PatrickJMT 

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\(\displaystyle{ \ln(3x^2+9x5) }\)
Problem Statement 

Calculate the derivative of this function and give your final answer in completely factored form. \(\displaystyle{ \ln(3x^2+9x5) }\)
Final Answer 

\(\displaystyle{ \frac{3(2x+3)}{3x^2+9x5} }\)
Problem Statement 

Calculate the derivative of this function and give your final answer in completely factored form. \(\displaystyle{ \ln(3x^2+9x5) }\)
Solution 

\(\displaystyle{ \frac{d}{dx}[ \ln(3x^2+9x5) ] }\) 
Apply the chain rule. 
\(\displaystyle{ \frac{1}{3x^2+9x5} \cdot \frac{d}{dx}[3x^2+9x5] }\) 
\(\displaystyle{ \frac{1}{3x^2+9x5} \cdot (6x+9) }\) 
\(\displaystyle{ \frac{3(2x+3)}{3x^2+9x5} }\) 
Using the substitution \( u = 3x^2+9x5 \) this could have solved more explicitly.
\(\displaystyle{ \frac{d}{dx}[ \ln(3x^2+9x5) ] }\) 
\(\displaystyle{ \frac{d}{du}[\ln(u)] \cdot \frac{d}{dx}[3x^2+9x5] }\) 
\(\displaystyle{ \frac{1}{u} \cdot (6x+9) }\) 
\(\displaystyle{ \frac{1}{3x^2+9x5} \cdot [ 3(2x+3) ] }\) 
Final Answer 

\(\displaystyle{ \frac{3(2x+3)}{3x^2+9x5} }\)
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\(h(x)=\ln(x^2+3x+4)\)
Problem Statement 

Calculate the derivative of \(h(x)=\ln(x^2+3x+4)\)
Solution 

video by PatrickJMT 

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\(f(x)=\ln(5x^2+2x7)\)
Problem Statement 

Calculate the derivative of \(f(x)=\ln(5x^2+2x7)\)
Solution 

video by MathTV 

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\(g(x)=\log_4(x^3+8x)\)
Problem Statement 

Calculate the derivative of \(g(x)=\log_4(x^3+8x)\)
Solution 

video by PatrickJMT 

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\(\displaystyle{f(x)=\ln\sqrt[3]{x^3x}}\)
Problem Statement 

Calculate the derivative of \(\displaystyle{f(x)=\ln\sqrt[3]{x^3x}}\)
Final Answer 

\(\displaystyle{ \frac{3x^21}{3x(x^21)} }\)
Problem Statement 

Calculate the derivative of \(\displaystyle{f(x)=\ln\sqrt[3]{x^3x}}\)
Solution 

video by Krista King Math 

Final Answer 

\(\displaystyle{ \frac{3x^21}{3x(x^21)} }\)
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\(\displaystyle{f(x)=\ln\left(x\sqrt{x^2+1}\right)}\)
Problem Statement 

Calculate the derivative of \(\displaystyle{f(x)=\ln\left(x\sqrt{x^2+1}\right)}\)
Solution 

video by Krista King Math 

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Calculate the first three derivatives of \(\displaystyle{ f(x)=\ln(2+3x) }\) and give your answers in completely factored form.
Problem Statement 

Calculate the first three derivatives of \(\displaystyle{ f(x)=\ln(2+3x) }\) and give your answers in completely factored form.
Solution 

First Derivative
\(\displaystyle{ f'(x) = \frac{d}{dx}[ \ln(2+3x) ] }\) 
\(\displaystyle{ \frac{1}{2+3x} \frac{d}{dx}[2+3x] }\) 
\(\displaystyle{ \frac{1}{2+3x} (3) = \frac{3}{2+3x} }\) 
Second Derivative
We can use the either quotient rule or product rule here. The product rule is the easiest here because of the constant in the numerator but we need to rewrite the first derivative as \(\displaystyle{ \frac{3}{2+3x} = 3(2+3x)^{1} }\).
\(\displaystyle{ f''(x) = \frac{d}{dx} \left[ 3(2+3x)^{1} \right] }\) 
\(\displaystyle{3(1)(2+3x)^{2} \frac{d}{dx}[2+3x] }\) 
\(\displaystyle{3(2+3x)^{2} (3) = \frac{9}{(2+3x)^{2}} }\) 
Third Derivative
Again, we can use either the quotient rule or product rule and we choose the product rule.
\( f^{(3)}(x) = \displaystyle{ \frac{d}{dx}\left[ 9(2+3x)^{2} \right]} \) 
\(\displaystyle{9(2)(2+3x)^{3}\frac{d}{dx}[2+3x]} \) 
\( 18(2+3x)^{3} (3) \) 
\(\displaystyle{\frac{54}{(2+3x)^3} }\) 
Final Answers 

\(\displaystyle{f'(x) = \frac{3}{2+3x}}\) 
\(\displaystyle{f''(x) = \frac{9}{(2+3x)^{2}}}\) 
\(\displaystyle{ f^{(3)}(x)=\frac{54}{(2+3x)^3} }\) 
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Intermediate 

\(\displaystyle{f(x)=\ln\left[\frac{(2x+1)^5}{\sqrt{x^2+1}}\right]}\)
Problem Statement 

Calculate the derivative of \(\displaystyle{f(x)=\ln\left[\frac{(2x+1)^5}{\sqrt{x^2+1}}\right]}\)
Solution 

video by Krista King Math 

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\(\displaystyle{ f(x)=\ln\left[ \frac{(2x+1)^3}{(3x1)^4} \right] }\)
Problem Statement 

Calculate the derivative of \(\displaystyle{ f(x)=\ln\left[ \frac{(2x+1)^3}{(3x1)^4} \right] }\)
Solution 

video by PatrickJMT 

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\(y=\sqrt[3]{\log_7(x)}\)
Problem Statement 

Calculate the derivative of \(y=\sqrt[3]{\log_7(x)}\)
Solution 

video by PatrickJMT 

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\(y=\ln(x^4\sin x)\)
Problem Statement 

Calculate the derivative of \(y=\ln(x^4\sin x)\)
Solution 

video by PatrickJMT 

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\(y=[\log_4(1+e^x)]^2\)
Problem Statement 

Calculate the derivative of \(y=[\log_4(1+e^x)]^2\)
Solution 

video by PatrickJMT 

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\(p(x)=\ln\left[ x^2 \cdot \sqrt{x^3+3x} \cdot (x+2)^4 \right] \)
Problem Statement 

Calculate the derivative of \(p(x)=\ln\left[ x^2 \cdot \sqrt{x^3+3x} \cdot (x+2)^4 \right] \)
Solution 

video by PatrickJMT 

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You CAN Ace Calculus
For the basic logarithm 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.