Calculus

Differentiation of Functions

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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.

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

Click or tap a problem to see the solution.

Example 1

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

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 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.

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