# Derivatives of Logarithmic Functions

• On the page Definition of the Derivative, we have found the expression for the derivative of the natural logarithm function $$y = \ln x:$$

$\left( {\ln x} \right)^\prime = \frac{1}{x}.$

Now we consider the logarithmic function with arbitrary base and obtain a formula for its derivative.

So, let’s take the logarithmic function $$y = {\log _a}x,$$ where the base $$a$$ is greater than zero and not equal to $$1:$$ $$a \gt 0$$, $$a \ne 1$$. According to the definition of the derivative, we give an increment $$\Delta x \gt 0$$ to the independent variable $$x$$ assuming that $$x + \Delta x \gt 0$$. The logarithmic function will increment, respectively, by the value of $$\Delta y$$ where

${\Delta y }={ {\log _a}\left( {x + \Delta x} \right) – {\log _a}x.}$

Divide both sides by $$\Delta x:$$

${\frac{{\Delta y}}{{\Delta x}} }={ \frac{1}{{\Delta x}}\left[ {{{\log }_a}\left( {x + \Delta x} \right) – {{\log }_a}x} \right] } = {\frac{1}{{\Delta x}}{\log _a}\frac{{x + \Delta x}}{x} } = {\frac{1}{{\Delta x}}{\log _a}\left( {1 + \frac{{\Delta x}}{x}} \right).}$

Denote $${\large\frac{{\Delta x}}{x}\normalsize} = {\large\frac{1}{n}\normalsize}$$. Then the last relation can be rewritten as

${\frac{{\Delta y}}{{\Delta x}} }={ \frac{1}{{\Delta x}}{\log _a}\left( {1 + \frac{{\Delta x}}{x}} \right) } = {\frac{1}{x} \cdot n\,{\log _a}\left( {1 + \frac{1}{n}} \right).}$

Using the power property for logarithms, we obtain:

${\frac{{\Delta y}}{{\Delta x}} }={ \frac{1}{x}{\log _a}{\left( {1 + \frac{1}{n}} \right)^n}.}$

Supposing that $$\Delta x \to 0$$ (in this case $$n \to \infty$$), we find the limit of the ratio of the increments, i.e. the derivative of the logarithmic function:

${\lim\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} } = {\lim\limits_{n \to \infty } \left[ {\frac{1}{x}{{\log }_a}{{\left( {1 + \frac{1}{n}} \right)}^n}} \right] } = {\frac{1}{x}{\log _a}\left[ {\lim\limits_{n \to \infty } {{\left( {1 + \frac{1}{n}} \right)}^n}} \right].}$

Here we used the property of the limit of a composite function given that the logarithmic function is continuous. The limit in the square brackets converges to the famous trancendential number $$e$$, which is approximately equal to $$2.718281828\ldots:$$

${\lim\limits_{n \to \infty } {\left( {1 + \frac{1}{n}} \right)^n} }={ e }\approx {2.718281828459 \ldots }$

Consequently, the derivative of the logarithmic function has the form

${\left( {{{\log }_a}x} \right)^\prime } = \frac{1}{x}{\log _a}e.$

By the change-of-base formula for logarithms, we have:

${{\log _a}e = \frac{{\ln e}}{{\ln a}} }={ \frac{1}{{\ln a}}.}$

Thus,

${y’\left( x \right) = {\left( {{{\log }_a}x} \right)^\prime } }={ \frac{1}{{x\ln a}}.}$

If $$a = e$$, we obtain the natural logarithm the derivative of which is expressed by the formula $${\left( {\ln x} \right)^\prime } = {\large\frac{1}{x}\normalsize}.$$

We note another important special case − the derivative of the common logarithm (to base $$10$$):

${{\left( {\log _{10}x} \right)^\prime } }={ \frac{{\log _{10}e}}{x} = \frac{M}{x},}$

where the number $$M$$ is equal to $$M = {\log _{10}}e \approx 0.43429 \ldots$$

Note that we derived the formula $$\left( {{{\log }_a}x} \right)^\prime = \large{\frac{1}{{x\ln a}}}\normalsize$$ from first principles – using the limit definition of the derivative. As the logarithmic function with base $$a$$ $$\left({a \gt 0}\right.$$, $$\left.{a \ne 1}\right)$$ and exponential function with the same base form a pair of mutually inverse functions, the derivative of the logarithmic function can also be found using the inverse function theorem.

Suppose we are given a pair of mutually inverse functions $$y = f\left( x \right) = {\log_a}x$$ and $$x = \varphi \left( y \right) = {a^y}.$$ Then

${\left( {{{\log }_a}x} \right)^\prime = f^\prime\left( x \right) }={ \frac{1}{{\varphi ^\prime\left( y \right)}} }={ \frac{1}{{\left( {{a^y}} \right)^\prime}} }={ \frac{1}{{{a^y}\ln a}} }={ \frac{1}{{{a^{{{\log }_a}x}}\ln a}} }={ \frac{1}{{x\ln a}}.}$

In the particular case $$a = e$$, the derivative is given by

$\left( {\ln x} \right)^\prime = \frac{1}{x}.$

In the examples below, determine the derivative of the given function.

• ## Solved Problems

Click a problem to see the solution.

### Example 1

$y = \frac{{\ln x}}{x}$

### Example 2

$y = x\ln x – x$

### Example 3

$y = x\ln {\frac{1}{x}}$

### Example 4

$y = \ln \left( {{x^2} – 2x} \right)$

### Example 5

$y = \frac{1}{{\ln x}}$

### Example 6

$y = \ln \left( {\sin x} \right)$

### Example 7

$y = {\log _2}\cos x$

### Example 8

$y = {\log _3}\frac{3}{x} + \frac{3}{x}$

### Example 9

$y = {\log _3}\left( {4{x^2}} \right)$

### Example 10

$y = {x^p}\ln x.$

### Example 11

$y = \ln \tan \frac{x}{2}$

### Example 12

$y = \ln \left( {x + \sqrt {{x^2} + {a^2}} } \right)$

### Example 13

$y = \ln {\frac{1}{{\sqrt {1 – {x^4}} }}}$

### Example 14

Compute the derivative of the function $$y = \large{\frac{{{x^2}}}{{\ln x}}}\normalsize$$ at $$x = e.$$

### Example 15

$y = \ln \sqrt {\frac{{1 – x}}{{1 + x}}}$

### Example 16

$y = \ln \left( {\arccos \frac{1}{x}} \right)$

### Example 17

$y = \ln \left( {\ln \cot x} \right)$

### Example 18

$y = \frac{{{{\log }_2}\left( {{x^2}} \right)}}{{{x^2}}}$

### Example 19

$y = {\log_2}x \cdot {\log _3}x$

### Example 20

$y = \ln \left( {\tan x + \sec x} \right)$

### Example 21

$y = {\log_x}2.$

### Example 22

Calculate the derivative of the function $$y = {\log_2}x\ln \left( {2x} \right)$$ at $$x = 1.$$

### Example 1.

$y = \frac{{\ln x}}{x}$

Solution.

Differentiate using the quotient rule:

${y’\left( x \right) = {\left( {\frac{{\ln x}}{x}} \right)^\prime } } = {\frac{{{{\left( {\ln x} \right)}^\prime } \cdot x – \ln x \cdot x’}}{{{x^2}}} } = {\frac{{\frac{1}{x} \cdot x – \ln x \cdot 1}}{{{x^2}}} } = {\frac{{1 – \ln x}}{{{x^2}}},}$

where $$x \gt 0.$$

### Example 2.

$y = x\ln x – x$

Solution.

Using the product and difference rules, we have

$\require{cancel} {y’\left( x \right) = {\left[ {x\ln x – x} \right]^\prime } } = {{\left( {x\ln x} \right)^\prime } – x’ } = {x’\ln x + x{\left( {\ln x} \right)^\prime } – x’ } = {1 \cdot \ln x + x \cdot \frac{1}{x} – 1 } = {\ln x + \cancel{1} – \cancel{1} = \ln x\;\;}\kern-0.3pt{\left( {x \gt 0} \right).}$

### Example 3.

$y = x\ln {\frac{1}{x}}$

Solution.

Using the product rule, the chain rule and the derivative of the natural logarithm, we have

$\cssId{element14}{y^\prime = \left( {x\ln \frac{1}{x}} \right)^\prime }={ x^\prime \cdot \ln \frac{1}{x} + x \cdot \left( {\ln \frac{1}{x}} \right)^\prime }={ 1 \cdot \ln \frac{1}{x} + x \cdot \frac{1}{{\frac{1}{x}}} \cdot \left( {\frac{1}{x}} \right)^\prime }={ \ln \frac{1}{x} + x \cdot x \cdot \left( { – \frac{1}{{{x^2}}}} \right) }={ \ln \frac{1}{x} – \frac{{\cancel{x^2}}}{{\cancel{x^2}}} }=\cssId{element15}{ \ln \frac{1}{x} – 1.}$

### Example 4.

$y = \ln \left( {{x^2} – 2x} \right)$

Solution.

${y^\prime = \left[ {\ln \left( {{x^2} – 2x} \right)} \right]^\prime }={ \frac{1}{{{x^2} – 2x}} \cdot \left( {{x^2} – 2x} \right)^\prime }={ \frac{{2x – 2}}{{{x^2} – 2x}}.}$

### Example 5.

$y = \frac{1}{{\ln x}}$

Solution.

By the power rule and the chain rule,

${y^\prime = \left( {\frac{1}{{\ln x}}} \right)^\prime }={ \left[ {{{\left( {\ln x} \right)}^{ – 1}}} \right]^\prime }={ – 1 \cdot {\left( {\ln x} \right)^{ – 2}} \cdot \left( {\ln x} \right)^\prime }={ – \frac{1}{{{{\ln }^2}x}} \cdot \frac{1}{x} }={ – \frac{1}{{x{{\ln }^2}x}}.}$

### Example 6.

$y = \ln \left( {\sin x} \right)$

Solution.

By the chain rule,

${y^\prime = \left( {\ln \left( {\sin x} \right)} \right)^\prime }={ \frac{1}{{\sin x}} \cdot \left( {\sin x} \right)^\prime }={ \frac{1}{{\sin x}} \cdot \cos x }={ \frac{{\cos x}}{{\sin x}} }={ \cot x.}$

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Problems 1-6
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Problems 7-22