Derivatives of Exponential Functions
On this page we'll consider how to differentiate exponential functions.
Exponential functions have the form f (x) = ax, 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.
Now 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
Example 1.
\[y = {\pi ^{\frac{1}{x}}}\]
Solution.
By the chain rule, we obtain:
\[y'\left( x \right) = \left( {{\pi ^{\frac{1}{x}}}} \right)^\prime = {\pi ^{\frac{1}{x}}} \cdot \ln \pi \cdot {\left( {\frac{1}{x}} \right)^\prime } = {\pi ^{\frac{1}{x}}} \cdot \ln \pi \cdot \left( { - \frac{1}{{{x^2}}}} \right) = - \frac{{{\pi ^{\frac{1}{x}}}\ln \pi }}{{{x^2}}}.\]
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
\[y^\prime = \left( {{4^{3{x^2}}}} \right)^\prime = {4^{3{x^2}}}\ln 4 \cdot \left( {3{x^2}} \right)^\prime = {4^{3{x^2}}}\ln 4 \cdot 6x = 6x{4^{3{x^2}}}\ln 4.\]
Example 6.
\[y = 3^{\frac{1}{x}}\]
Solution.
By the chain rule,
\[y^\prime = \left( {{3^{\frac{1}{x}}}} \right)^\prime = {3^{\frac{1}{x}}}\ln 3 \cdot \left( {\frac{1}{x}} \right)^\prime = {3^{\frac{1}{x}}}\ln 3 \cdot \left( { - \frac{1}{{{x^2}}}} \right) = - \frac{{{3^{\frac{1}{x}}}\ln 3}}{{{x^2}}}.\]
See more problems on Page 2.
