Calculus

Differentiation of Functions

Differentiation of Functions Logo

Derivatives of Exponential Functions

  • On this page we’ll consider how to differentiate exponential functions.

    Exponential functions have the form \(f\left( x \right) = {a^x},\) where \(a\) is the base. The base is always a positive number not equal to \(1.\)

    If the base is equal to the number \(e:\)

    \[{a = e \approx 2.718281828 \ldots ,}\]

    then the derivative is given by

    \[{\frac{d}{{dx}}\left( {{e^x}} \right) = \left( {{e^x}} \right)^\prime = {e^x}.}\]

    (This formula is proved on the page Definition of the Derivative.)

    The function \(y = {e^x}\) is often referred to as simply the exponential function.

    Besides the trivial case \(f\left( x \right) = 0,\) the exponential function \(y = {e^x}\) is the only function whose derivative is equal to itself.

    No we consider the exponential function \(y = {a^x}\) with arbitrary base \(a\) \(\left( {a \gt 0, a \ne 1} \right)\) and find an expression for its derivative.

    As \(a = {e^{\ln a}},\) then

    \[{a^x} = {\left( {{e^{\ln a}}} \right)^x} = {e^{x\ln a}}.\]

    Using the chain rule, we have

    \[{\left( {{a^x}} \right)^\prime = \left( {{e^{x\ln a}}} \right)^\prime}={ {e^{x\ln a}} \cdot \left( {x\ln a} \right)^\prime}={ {e^{x\ln a}} \cdot \ln a }={ {a^x} \cdot \ln a.}\]

    Thus

    \[\left( {{a^x}} \right)^\prime = {a^x}\ln a.\]

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


  • Solved Problems

    Click a problem to see the solution.

    Example 1

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

    Example 2

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

    Example 3

    \[y = {2^{\sqrt x }}\left( {x \gt 0} \right)\]

    Example 4

    \[y = {e^{ – {x^3}}}\]

    Example 5

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

    Example 6

    \[y = 3^{\frac{1}{x}}\]

    Example 7

    \[y = {2^{ – {x^2}}}\]

    Example 8

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

    Example 9

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

    Example 10

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

    Example 11

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

    Example 12

    \[y =\frac{{{e^x} – 1}}{{{e^x} + 1}}\]

    Example 13

    \[y = {e^x}\left( {{x^2} – 2x + 2} \right)\]

    Example 14

    \[y = \frac{{{e^x} – {e^{ – x}}}}{{{e^x} + {e^{ – x}}}}\]

    Example 15

    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 16

    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 1.

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

    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 2.

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

    Solution.

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

    Example 3.

    \[y = {2^{\sqrt x }}\left( {x \gt 0} \right)\]

    Solution.

    Using the chain rule, we get

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

    Example 4.

    \[y = {e^{ – {x^3}}}\]

    Solution.

    By the chain rule,

    \[{y^\prime = \left( {{e^{ – {x^3}}}} \right)^\prime }={ {e^{ – {x^3}}} \cdot \left( { – {x^3}} \right)^\prime }={ {e^{ – {x^3}}} \cdot \left( { – 3{x^2}} \right) }={ – 3{x^2}{e^{ – {x^3}}}.}\]

    Example 5.

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

    Solution.

    Using the chain rule, we have

    \[{y^\prime = \left( {{4^{3{x^2}}}} \right)^\prime }={ {4^{3{x^2}}}\ln 4 \cdot \left( {3{x^2}} \right)^\prime }={ {4^{3{x^2}}}\ln 4 \cdot 6x }={ 6x{4^{3{x^2}}}\ln 4.}\]

    Page 1
    Problems 1-5
    Page 2
    Problems 6-16