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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 A01

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

answer

solution

Practice A02

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

answer

solution

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

\(\displaystyle{y=\frac{x^4\sqrt{x-3}}{(x+1)^3}}\)

solution

Practice A10

\(\displaystyle{y=\sqrt{\frac{1-x}{1+x}}}\)

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 A14

\(\displaystyle{y=\frac{2(x^2+1)}{\sqrt{\cos(2x)}}}\)

answer

solution

Practice A15

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

solution

Practice A16

\(\displaystyle{f(x)=(2x-3)^2(5x^2+2)^3}\)

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 \(\displaystyle{f(x)=\frac{x^2(x+2)^4}{(2x^2-1)^3}}\), find \(f'(x)\) and \(f'(4)\)

solution

Practice B04

\(\displaystyle{y=\sqrt[3]{\frac{x(x^2+1)^4}{x^3-2}}}\)

solution


Level C - Advanced

Practice C01

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

solution

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