# Derivatives of 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 or tap 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

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

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Problems 1-5
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Problems 6-16