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WikiBooks  Derivatives of Exponential and Logarithm Functions 
Logarithmic Differentiation 

This topic is usually found in the section discussing implicit differentiation and sometimes instructors do not make a distinction between the two. But logarithmic differentiation is a very specific technique and often uses implicit differentiation along the way. 

There are two main types of equations that you will use logarithmic differentiation on
1. equations where you have a variable in an exponent
2. equations that are quite complicated and can be simplified using logarithms.
In both cases, we introduce logarithms into the equation that may not have been there before, apply some simple rules and then take the derivative. Let's look at each case.
Variables In The Exponent 

Remember that you can use the power rule on \(x^2\) but you can't use the power rule on \(2^x\) or \(y^x\). [ why? ] So, what we do is introduce a natural log into the equation, without changing the problem of course. The goal is to bring the exponent down so that we can take the derivative of it.
These are the rules we use 

1. \( \ln(x^y) = y~ \ln(x) \) 
2. \( e^{\ln(z)} = z \) 
Notice the first rule brings the exponent down in front of the natural log term, in which case, we can use the product rule to take the derivative. The second rule is usually used to reverse the process after taking the derivative.
Let's look at an example. One of the practice problems shows how to calculate the derivative of \( y = x^x \). To do this one, we need to bring the x in the exponent down (since we can't use the power rule). To accomplish this, we take the natural log of both sides, like this:
\(\displaystyle{
\begin{array}{rcl}
y & =& x^x \\
\ln(y) & = & \ln(x^x) \\
\ln(y) & = & x \ln(x)
\end{array}
}\)
Now we can take the derivative of this last equation using chain rule and the product rule. [ For a complete solution, see practice problem A01. ]
Simplifying 

In some cases, we could use the product and/or quotient rules to take a derivative but, using logarithmic differentiation, the derivative would be much easier to find. One of the practice problems is to take the derivative of \(\displaystyle{ y = \frac{(\sin(x))^2(x^3+1)^4}{(x+3)^8} }\). We could use the product and quotient rules here but, if we take the logarithm of both sides, simplify, take the derivative, then convert back, it is much easier. [ see practice problem B01 for how to find the derivative \( dy/dx \) ]
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Practice Problems 

Instructions   Unless otherwise instructed, calculate \( dy/dx \) of the following equations using logarithmic differentiation. Give your final answers in exact, completely factored form.
Level A  Basic 
Practice A03  

\(\displaystyle{y=(\ln x)^x}\)  
solution 
Practice A04  

\(\displaystyle{y=(x^2)^{\sin(x)}}\)  
solution 
Practice A05  

\(\displaystyle{y=5^x}\)  
solution 
Practice A06  

\(\displaystyle{y = x^{\sin(x)}}\)  
solution 
Practice A07  

\(\displaystyle{y=\frac{(x+2)^2}{\sqrt{x^2+1}}}\)  
solution 
Practice A08  

for \(\displaystyle{f(x)=x^{2x}}\), find \(f'(x)\) and \(f'(1)\)  
solution 
Practice A09  

\(\displaystyle{y=\frac{x^4\sqrt{x3}}{(x+1)^3}}\)  
solution 
Practice A10  

\(\displaystyle{y=\sqrt{\frac{1x}{1+x}}}\)  
solution 
Practice A11  

\(\displaystyle{y=x^{x^2}}\)  
solution 
Practice A12  

\(\displaystyle{y=(\sin(x))^x}\)  
solution 
Practice A13  

\(\displaystyle{y=(\cos x)^{\tan x}}\)  
solution 
Practice A15  

\(\displaystyle{y=(\sin \theta)^{\sqrt{\theta}}}\)  
solution 
Practice A16  

\(\displaystyle{f(x)=(2x3)^2(5x^2+2)^3}\)  
solution 
Level B  Intermediate 
Practice B01  

\(\displaystyle{y=\frac{(\sin(x))^2(x^3+1)^4}{(x+3)^8}}\)  
solution 
Practice B02  

\(\displaystyle{y=\sqrt{x}e^{x^2}(x^2+1)^{10}}\)  
solution 
Practice B03  

for \(\displaystyle{f(x)=\frac{x^2(x+2)^4}{(2x^21)^3}}\), find \(f'(x)\) and \(f'(4)\)  
solution 
Practice B04  

\(\displaystyle{y=\sqrt[3]{\frac{x(x^2+1)^4}{x^32}}}\)  
solution 
Level C  Advanced 
Practice C01  

\(\displaystyle{y=x^{(x^x)}}\)  
solution 