Using Logarithmic Differentiation to Compute Derivatives

Instructor: Gerald Lemay

Gerald has taught engineering, math and science and has a doctorate in electrical engineering.

In this lesson, we will explore logarithmic differentiation and show how this technique relates to certain types of functions. Using two examples, we will learn how to compute derivatives using this differentiation method.

Logarithmic Differentiation

Oversleeping and skipping class, Fred checks in with friends to see what he missed. ''Logarithmic differentiation,'' they say. Sounds like what a lumberjack needs to know when selecting trees to cut. Before Fred heads out to the hardware store to buy a chainsaw, let's point him to this lesson.

Using logarithms, we will be able to find the derivative of functions that have x in the exponent. Our examples are y = xx and y = (4x3 - 2)1/x. See the x in the exponent? Anyone see Fred?

Natural Logarithms

There goes Fred heading out into nature in search of logs to differentiate. Actually, he is somewhat correct because the log he needs is the natural logarithm; the logarithm with base e. This log of x, written as ln x, has a simple derivative equal to 1/x. Also, like all logarithms, it can be used to free up an exponent: ln xa is a ln x. The exponent a is brought out in front.

The standard rules for differentiating like the power rule and the product rule won't work directly on functions with exponents in x.

Example 1: Differentiate y = xx for x > 0

Before differentiating y = xx, let's plot this function.

y = x^x

In this example we will

  • differentiate the function to get y '
  • confirm the y ' result by plotting a tangent line



Step 1: Take the natural log.


The right-hand side, ln xx, becomes x ln x.


Step 2: Differentiate.

On the left-hand side, the derivative of ln y is 1/y times y '. On the right-hand side, the derivative of x ln x is found using the product rule:

(derivative of x)(ln x) + x(derivative of ln x)

= (1)(ln x) + x(1/x)

= ln x + x/x

= ln x + 1



Step 3: Solve for y '.

Multiplying both sides by y:


Step 4: Substitute for y on the right-hand side.

Replace y with xx:


We have a result for the derivative but how can we verify this answer?

The derivative is the slope of the tangent line at a point. Let's use our y ' result to find and plot the tangent line at x = 1.

First, evaluate the derivative at x = 1:

y ' = xx (1 + ln x) becomes

y ' = 11 (1 + ln 1) or

y ' = 1 (1 + 0) = 1

The equation for a line has the form y = mx + b. We have the slope m = 1. To find b, determine the y value where the slope touches the curve. From

y = xx, let x = 1:

y = 11 = 1.

The equation for the tangent line must satisfy y = mx + b where y = 1 and m = 1. Thus,

1 = 1(x) + b where x = 1. Thus, b = 0. The equation for the tangent line at x = 1 is y = mx + b = 1(x) + 0 = x. Thus, the tangent line at x = 1 is y = x.

Plotting this tangent line with the curve looks good:

Tangent line at x=1

Speaking of tangents, Fred has been spotted carrying a field guide for identifying trees. How can we get him on the right track? Maybe, another example.

Example 2: Differentiate y = (4x3 - 2)1/x for x > 0

There's an x in the exponent again. Time for logarithmic differentiation.

y=(4x^3 -2)^(1/x)
(4x^3 -2)^(1/x)

In this example we will

  • differentiate the function
  • plot a tangent line

Step 1: Take the natural log.

From y = (4x3 - 2)1/x, take the natural log of both sides:

ln y = ln (4x3 - 2)1/x


ln y = (1/x) ln (4x3 - 2)

Step 2: Differentiate.

The left-hand side is (y ')/y as before.

The right-hand side is differentiated using the product rule. So far, we have:


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