# 6.3: Derivatives of Exponential Functions

Exponential functions, and the rate of change, are used to model many real-world situations such as population growth, radioactive half-life decay, attenuation of electromagnetic signals in media, and financial transactions. Do you know how to write general exponential equations for the growth of a population that doubles every 5 years, and its rate of change?

## Derivatives of Exponential Functions

An exponential function $$f(x) \nonumber$$ has the form:

$f(x) = b^x \nonumber$

where b is called the base and is a positive, real number.

The figure below shows a few exponential function graphs for 0<b≤10. It is very clear that the sign of the derivative of an exponential depends on the value of b. If 0<b<1, the value of the derivative of the function (slope of the tangent line) will be negative because the function is always decreasing as x increases. For b>1, the derivative of the function will always be positive because the function increases as x increases.

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But, what is the derivative of an exponential function? We can take the following steps to find an expression for $$\frac{d}{dx}[bx] \nonumber$$ by using the definition of the derivative:

$$\displaystyle \frac{d}{dx}[bx]= \lim_{h \to 0} \frac{b^{x+h}−b^x}{h} \nonumber$$.... Limit definition of the derivative

$$= \lim_{h \to 0} \frac{b^xb^h−b^x}{h} \nonumber$$....Exponent property

$$= \lim_{h \to 0} \frac{b^h−1}{h}⋅b^x \nonumber$$....Factoring

$$= ( \lim_{h→0} \frac{b^h−1}{h})⋅bx \nonumber$$....Limit of a product property

The result above shows $$\frac{d}{dx}[bx] \nonumber$$ depends on the product of $$\lim_{h→0} \frac{b^h−1}{h} \nonumber$$ and the original function. But what is $$\lim_{h→0} \frac{b^h−1}{h} \nonumber$$? There are a number of ways to evaluate this limit, but for now let’s take a quick look at the behavior of $$\frac{b^h−1}{h} \nonumber$$. This function is graphed below for a number of values of b, and the limit at $$h=0 \nonumber$$ is indicated by the points A−E.

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While it is not at all obvious:$$\lim_{h→0} \frac{b^h−1}{h} = ln(b) \nonumber$$. Remember the natural logarithm function $$y=lnx \nonumber$$ on your calculator? The natural logarithm is the general logarithm function with base $$b=e=2.71828 \nonumber$$...

Given the exponential function $$f(x)=b^x \nonumber$$, where the base b is a positive, real number, then the general representation of the derivative of an exponential function is:

$\frac{d}{dx}[bx]=lnb⋅b^x \nonumber$

Adding the Chain Rule to the definition, given the exponential function f(x)=bu, where u=g(x) and g(x) is a differentiable function, then:

$\frac{d}{dx}[bu]=(lnb⋅b^u) \frac{du}{dx} \nonumber$

## Examples

### Example 1

Earlier, you were asked what the general exponential equation for the growth of a population that doubles every 5 years is.

A population $$P(t) \nonumber$$ that doubles every 5 years could be modeled as $$P(t)=P_02^{\frac{t}{5}} \nonumber$$, where the variable t represents number of years since the population was at a level of $$P_0 \nonumber] \nonumber$$. Were you able to determine that the rate of change of $$P(t) \nonumber$$ is $$P′(t) = \frac{P_0 ln2}{5}⋅2^{\frac{t}{5}} \nonumber$$?

### Example 2

Given $$y=500⋅0.7^x \nonumber$$, what is $$\frac{dy}{dx} \nonumber$$?

$\frac{dy}{dx}=\frac{d}{dx}[500⋅0.7^x] \nonumber$

$= 500 \frac{d}{dx}[0.7^x] \nonumber$

$$= 500[ln(0.7)⋅0.7^x] \nonumber$$....Use your calculator to find ln(0.7)

$=−178.3⋅0.7^x \nonumber$

Hence, $$\frac{dy}{dx} =−178.3⋅0.7^x \nonumber$$, and as expected, the slopes of all tangent lines are negative.

There is an important special case that you must know about

### Example 3

Given $$y=500⋅e^x \nonumber$$, what is $$\frac{dy}{dx} \nonumber$$?

$\frac{dy}{dx}= \frac{d}{dx}[500⋅ex] \nonumber$

$= 500 \frac{d}{dx}[ex] \nonumber$

$$\frac{dy}{dx}=500[ln(e)⋅ex] \nonumber$$....Use your calculator to find ln(e)

$= \frac{d}{dx}[500⋅ex] \nonumber$

$=500[1⋅ex] \nonumber$

$=500⋅ex \nonumber$

Hence, $$\frac{dy}{dx}=500⋅ex \nonumber$$, and this is just the original function. This exponential function, with base e, is special: the rate of change (or slope of the tanget line) at any point is equal to the value of the function at that point.

### Example 4

Given $$y=10⋅2.5−3x^2 \nonumber$$, what is $$\frac{dy}{dx} \nonumber$$?

$\frac{dy}{dx} = \frac{d}{dx}[10⋅2.5−3x2] \nonumber$

$=10 \frac{d}{dx}[2.5−3x2] \nonumber$

$=10⋅ln(2.5)⋅2.5−3x2⋅ddx[−3x^2] \nonumber$

$=10⋅(0.9162)⋅2.5−3x^2⋅[−6x] \nonumber$

$=−55x⋅2.5−3x^2 \nonumber$

Therefore, $$\frac{dy}{dx} =−55x⋅2.5−3x^2 \nonumber$$

### Example 5

Given $$y=500⋅e−2x⋅cos(5πx) \nonumber$$, what is $$\frac{dy}{dx} \nonumber$$?

$\frac{dy}{dx}= \frac{d}{dx}[500⋅e−2x⋅cos(5πx)] \nonumber$

$$=500⋅[ \frac{d}{dx}(e−2x)⋅cos(5πx)+e−2x⋅ \frac{d}{dx}(cos(5πx))] \nonumber$$....Product Rule

$$=500⋅[ln(e)⋅e−2x \frac{d}{dx}(−2x)⋅cos(5πx)+e−2x⋅(−sin(5πx)⋅ \frac{d}{dx}(5πx))] \nonumber$$....Use Chain Rule

$$=500⋅[(1)⋅e−2x⋅(−2)⋅cos(5πx)+e−2x⋅(−sin(5πx)⋅5π)] \nonumber$$....Simplify.

$$=−500⋅e−2x[2⋅cos(5πx)+5πsin(5πx)] \nonumber$$...Simplify

Therefore, $$\frac{dy}{dx} =−500⋅e−2x[2⋅cos(5πx)+5πsin(5πx)] \nonumber$$.

## Review

For #1-14, find the derivative.

1. $$y=7^x \nonumber$$
2. $$y=3^{2x} \nonumber$$
3. $$y=5^x−3x^2 \nonumber$$
4. $$y=2^{x^2} \nonumber$$
5. $$y=e^{x^2} \nonumber$$
6. $$f(x)= \frac{1}{ \sqrt{πσ}}e^{−αk(x−x0)^2} \nonumber$$ where σ, α, x0, and k are constants and σ≠0.
7. $$y=e^{6x} \nonumber$$
8. $$y=e^{3x^3}−2x^2+6 \nonumber$$
9. $$y=\frac{e^x−e−x}{e^x+e−x} \nonumber$$
10. $$y=cos(e^x) \nonumber$$
11. $$y=e^{−x}3^x \nonumber$$
12. $$y=3^{−x^2+2x+1} \nonumber$$
13. $$y=2^x3^x \nonumber$$
14. $$y=e^{−x}sinx \nonumber$$
15. Find an equation of the tangent line to $$f(x)=x^3+2e^x \nonumber$$ at the point (0, 2).

Exponential Function An exponential function is a function whose variable is in the exponent. The general form is $$y=a⋅b^{x−h}+k \nonumber$$.