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17Calculus - Logarithmic Differentiation

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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 \(y^x\).

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 890.]

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 892 for how to find the derivative \( dy/dx \)]

Practice

Basic Problems

\( y=5^x \)

Problem Statement

\( y=5^x \)

Solution

1543 video

video by MathTV

close solution
\( y = x^x \)

Problem Statement

\( y = x^x \)

Final Answer

\( d[x^x]/dx = x^x(1+\ln(x)) \)

Problem Statement

\( y = x^x \)

Solution

\( \ln(y) = \ln(x^x) \)

\( \ln(y) = x \ln(x) \)

\(\displaystyle{ \frac{d}{dx}[\ln(y)] = \frac{d}{dx}[x \ln(x)] }\)

\( (1/y)(dy/dx) = x(1/x) + \ln(x)(1) \)

\( dy/dx = y(1+\ln(x)) \)

\( dy/dx = x^x(1+\ln(x)) \)

Notice we could not use the power rule since the exponent is not a rational number.
Here is a video solution.

890 video

video by Khan Academy

Final Answer

\( d[x^x]/dx = x^x(1+\ln(x)) \)

close solution
For \( f(x)=x^{2x} \), determine \(f'(x)\) and \(f'(1)\).

Problem Statement

For \( f(x)=x^{2x} \), determine \(f'(x)\) and \(f'(1)\).

Solution

1546 video

video by MIP4U

close solution
\( y = x^{e^x} \)

Problem Statement

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

Final Answer

\( dy/dx=x^{e^x-1}e^x(1+x\ln(x)) \)

Problem Statement

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

Solution

Notice we cannot use the power rule since the exponent, \(e^x\), is not a rational number. So to get the factor out of the exponent, we use logarithms.

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

\( \ln(y) = e^x \ln(x) \)

\(\displaystyle{ \frac{1}{y} \cdot \frac{dy}{dx} = e^x (1/x) + \ln(x) e^x }\)

\(\displaystyle{ \frac{dy}{dx} = ye^x(1/x+\ln(x)) }\)

\(\displaystyle{ \frac{dy}{dx} = \frac{ye^x}{x}(1+x \ln(x)) }\)

\(\displaystyle{ \frac{dy}{dx} = \frac{x^{e^x}e^x}{x}(1+x \ln(x)) }\)

\(\displaystyle{ \frac{dy}{dx} = x^{e^x-1}e^x(1+x \ln(x)) }\)

Final Answer

\( dy/dx=x^{e^x-1}e^x(1+x\ln(x)) \)

close solution
\( y=(\ln x)^x \)

Problem Statement

\( y=(\ln x)^x \)

Solution

894 video

video by PatrickJMT

close solution
\( y = x^{\sin(x)} \)

Problem Statement

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

Solution

1544 video

video by MathTV

close solution
\(y=(x^2)^{\sin(x)}\)

Problem Statement

\(y=(x^2)^{\sin(x)}\)

Solution

895 video

video by PatrickJMT

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

Problem Statement

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

Solution

1545 video

video by MIP4U

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

Problem Statement

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

Solution

1548 video

video by MIP4U

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

Problem Statement

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

Solution

1550 video

close solution
\( y=x^{x^2} \)

Problem Statement

\( y=x^{x^2} \)

Solution

1552 video

close solution
\( y=(\sin(x))^x \)

Problem Statement

\( y=(\sin(x))^x \)

Solution

1553 video

close solution
\( y=(\cos x)^{\tan x} \)

Problem Statement

\( y=(\cos x)^{\tan x} \)

Solution

1554 video

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

Problem Statement

Use logarithmic differentiation to calculate \(dy/dx\) of \(\displaystyle{ y=\frac{2(x^2+1)}{\sqrt{\cos(2x)}} }\).

Final Answer

\(\displaystyle{ \frac{dy}{dx} = \left[ \frac{2x}{x^2+1}+\tan(2x) \right] \frac{2(x^2+1)}{\sqrt{\cos(2x)}} }\)

Problem Statement

Use logarithmic differentiation to calculate \(dy/dx\) of \(\displaystyle{ y=\frac{2(x^2+1)}{\sqrt{\cos(2x)}} }\).

Solution

There is a mistake in this video. When he first takes the derivative, he drops a factor of two from the first term on the right side of the equal sign. Actually, I would do one more simplification than he does before taking the derivative by expanding \(\ln [2(x^2+1)] = \ln 2 + \ln(x^2+1)\). Then when you take the derivative of \(\ln 2\) you get zero.

1555 video

video by SchoolOfChuck

Final Answer

\(\displaystyle{ \frac{dy}{dx} = \left[ \frac{2x}{x^2+1}+\tan(2x) \right] \frac{2(x^2+1)}{\sqrt{\cos(2x)}} }\)

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

Problem Statement

Use logarithmic differentiation to calculate \(dy/d\theta\) of \(y=(\sin \theta)^{\sqrt{\theta}}\).

Solution

1556 video

video by SchoolOfChuck

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

Problem Statement

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

Solution

1549 video

video by MIP4U

close solution

Intermediate Problems

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

Problem Statement

Use logarithmic differentiation to calculate the derivative of \(\displaystyle{ y = \frac{(\sin(x))^2(x^3+1)^4}{(x+3)^8} }\)

Solution

892 video

video by PatrickJMT

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

Problem Statement

Use logarithmic differentiation to calculate the derivative of \(\displaystyle{ y = \sqrt{x}e^{x^2}(x^2+1)^{10} }\)

Solution

893 video

video by PatrickJMT

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

Problem Statement

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

Solution

1547 video

video by MIP4U

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

Problem Statement

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

Solution

1551 video

close solution

Advanced Problems

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

Problem Statement

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

Solution

896 video

video by Khan Academy

close solution

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