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 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,

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

    Page 1
    Problems 1-6
    Page 2
    Problems 7-22