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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