Calculus

Differentiation of Functions

Differentiation of Functions Logo

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 or tap 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