Calculus

Differentiation of Functions

Derivatives of Exponential and Logarithmic Functions

Page 1
Problems 1-4
Page 2
Problems 5-18

On the Definition of the Derivative page we have derived formulas for the derivatives of the exponential function \(y = {e^x}\) and the natural logarithm function \(y = \ln x\).

Below we consider the exponential and logarithmic functions with arbitrary base and obtain expressions for their derivatives.

Derivative of Logarithmic Function

We start with the derivative of 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 \(M = {\log _{10}}e \approx 0.43429 \ldots \)

Derivative of Exponential Function

As the exponential function with base \(a\) \(\left({a \gt 0}\right.\), \(\left.{a \ne 1}\right)\) and logarithmic function with the same base form a pair of mutually inverse functions, the derivative of the exponential function can be found using the inverse function theorem.

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

Then
\[
{\left( {{a^x}} \right)^\prime = f’\left( x \right) }
= {\frac{1}{{\varphi’\left( y \right)}} }
= {\frac{1}{{{{\left( {{{\log }_a}y} \right)}^\prime }}} }
= {\frac{1}{{\frac{1}{{y\ln a}}}} }
= {y\ln a }
= {{a^x}\ln a.}
\] In the particular case \(a = e\), the derivative coincides with the function itself:
\[{{\left( {{e^x}} \right)^\prime } }={ {e^x}\ln e = {e^x}.}\]

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

Solved Problems

Click on problem description to see solution.

 Example 1

\[y = \frac{{\ln x}}{x}\]

 Example 2

\[y = {\pi ^{\large\frac{1}{x}\normalsize}}\]

 Example 3

\[y = x\ln x – x\]

 Example 4

\[y = {\log _2}\cos x\]

 Example 5

\[y = \sqrt {{2^x}} \]

 Example 6

\[y = {\log _3}\frac{3}{x} + \frac{3}{x}\]

 Example 7

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

 Example 8

\[y = \ln \tan \frac{x}{2}\]

 Example 9

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

 Example 10

\[y = {4^x} \cdot {3^{2x}}\]

 Example 11

\[y = \frac{{{x^2}}}{{{2^x}}}\]

 Example 12

\[y = \frac{{{x^n}}}{{{n^x}}}\;\left( {n \gt 0} \right)\]

 Example 13

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

 Example 14

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

 Example 15

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

 Example 16

Show that the function \(y = {e^x}\left( {\sin 2x + \cos 2x} \right)\) is a solution of the differential equation
\[y^{\prime\prime} – 2y’ + 5y = 0.\]

 Example 17

Show that the function \(y = {e^{ – 2x}}\left( {\sin 3x + \cos 3x} \right)\) is a solution of the differential equation
\[y^{\prime\prime} + 4y’ + 13y = 0.\]

 Example 18

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

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.

\[y = {\pi ^{\large\frac{1}{x}\normalsize}}\]

Solution.

By the chain rule, we obtain:
\[
{y’\left( x \right) = {\left( {{\pi ^{\large\frac{1}{x}\normalsize}}} \right)^\prime } }
= {{\pi ^{\large\frac{1}{x}\normalsize}} \cdot \ln \pi \cdot {\left( {\frac{1}{x}} \right)^\prime } }
= {{\pi ^{\large\frac{1}{x}\normalsize}} \cdot \ln \pi \cdot \left( { – \frac{1}{{{x^2}}}} \right) }
= { – \frac{{{\pi ^{\large\frac{1}{x}\normalsize}}\ln \pi }}{{{x^2}}}.}
\]

Example 3.

\[y = x\ln x – x\]

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 4.

\[y = {\log _2}\cos x\]

Solution.

Differentiating as a composite function, we can write:
\[
{y’\left( x \right) = {\left( {{{\log }_2}\cos x} \right)^\prime } }
= {\frac{1}{{\cos x \cdot \ln 2}} \cdot {\left( {\cos x} \right)^\prime } }
= {\frac{1}{{\cos x \cdot \ln 2}} \cdot \left( { – \sin x} \right) }
= { – \frac{{\sin x}}{{\cos x \cdot \ln 2}} }={ – \frac{{\tan x}}{{\ln 2}}.}
\] This function is defined only when
\[
{\cos x \gt 0,\;\;}\Rightarrow
{- \frac{\pi }{2} + 2\pi n \lt x \;}\kern0pt{\lt \frac{\pi }{2} + 2\pi n,\;\;}\kern-0.3pt{n \in \mathbb{Z}.}
\]

Page 1
Problems 1-4
Page 2
Problems 5-18