Differentiation of Exponential Functions

Neetesh Kumar | February 08, 2025 $\space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space$ Share this Page on:

• 1. Definition of Exponential Functions
• 2. Explanation & Intuition
• 3. Solved Examples
• 4. Conclusion

Exponential Function:
An exponential function is a function of the form: $f(x) = a \cdot b^x$, then

$f'(x) = \dfrac{d}{dx}\big(a.b^x\big) = a.b^x .\ln(b)$

where:

• $a$ is a constant coefficient.
• $b$ is a positive constant (base) greater than 0 and not equal to 1.
• $x$ is the variable (exponent).

The function represents exponential growth or decay, depending on the value of the base $b$.

Special Case:
When $b = e \approx 2.718$, the function is known as the natural exponential function and is denoted by:

$f(x) = a \cdot e^x$

$f'(x) = \dfrac{d}{dx}\big(a.e^x\big) = a.e^x .\ln(e) = a.e^x$

The derivative of an exponential function is particularly simple because the rate of change of exponential functions is proportional to their current value.

Example 1: Differentiating an Exponential Function

Differentiate $f(x) = 3e^{2x}$.

Solution:

Using the chain rule and knowing the derivative of $e^x$ is $e^x$, we differentiate:

$f'(x) = 3 \cdot e^{2x} \cdot \dfrac{d}{dx}(2x)$

$f'(x) = 3 \cdot e^{2x} \cdot 2$

$f'(x) = 6e^{2x}$

Thus, the final answer is: $\boxed{6e^{2x}}$

Example 2: Modeling Compound Interest

The amount $A$ of money in an account is given by the formula:

$A(t) = P e^{rt}$

where:

• $P$ is the principal,
• $r$ is the interest rate,
• $t$ is the time in years.

Find the rate of change of the amount after 5 years if $P = 1000$ and $r = 0.05$.

Solution:

To find the rate of change, differentiate $A(t) = P e^{rt}$ with respect to $t$:

$A'(t) = P \cdot e^{rt} \cdot \dfrac{d}{dt}(rt)$

$A'(t) = P \cdot e^{rt} \cdot r$

Substitute the given values $P = 1000$, $r = 0.05$, and $t = 5$:

$A'(5) = 1000 \cdot e^{0.05 \cdot 5} \cdot 0.05$

$A'(5) = 1000 \cdot e^{0.25} \cdot 0.05$

Using a calculator:

$A'(5) \approx 1000 \cdot 1.284 \cdot 0.05$

$A'(5) \approx 64.2$

Thus, the rate of change after 5 years is: $\boxed{64.2}$ dollars per year.

Example 3: Radioactive Decay

The amount of a radioactive substance is modeled by the function:

$N(t) = N_0 e^{-\lambda t}$

where:

• $N(t)$ is the amount remaining at time $t$,
• $N_0$ is the initial amount,
• $\lambda$ is the decay constant.

Find the rate of decay at $t = 10$ if $N_0 = 1000$ and $\lambda = 0.1$.

Solution:

The rate of decay is given by the derivative:

$N'(t) = N_0 e^{-\lambda t} \cdot (-\lambda)$

Substitute the given values $N_0 = 1000$, $\lambda = 0.1$, and $t = 10$:

$N'(10) = 1000 \cdot e^{-0.1 \cdot 10} \cdot (-0.1)$

$N'(10) = 1000 \cdot e^{-1} \cdot (-0.1)$

Using a calculator:

$N'(10) \approx 1000 \cdot 0.3679 \cdot (-0.1)$

$N'(10) \approx -36.79$

Thus, the rate of decay is: $\boxed{-36.79}$ units per time.

Example 4: Exponential Growth in Population

The population of a species is modeled by:

$P(t) = 500 e^{0.02t}$

where $t$ is the time in years. Find the rate of population growth at $t = 10$.

Solution:

The rate of growth is the derivative of $P(t)$:

$P'(t) = 500 e^{0.02t} \cdot 0.02$

Substitute $t = 10$:

$P'(10) = 500 e^{0.02 \cdot 10} \cdot 0.02$

$P'(10) = 500 e^{0.2} \cdot 0.02$

Using a calculator:

$P'(10) \approx 500 \cdot 1.221 \cdot 0.02$

$P'(10) \approx 12.21$

Thus, the rate of population growth at $t = 10$ is: $\boxed{12.21}$ individuals per year.