Differentiation of Logarithmic 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 Derivative of Logarithmic Functions
• 2. Formula for Derivative of Logarithmic Functions
• 3. Solved Examples
• 4. Conclusion

Derivative of Logarithmic Functions:
The derivative of a logarithmic function depends on the base of the logarithm. For a general logarithmic function, the derivative is given by:

$\boxed{\frac{d}{dx} \bigg[ \log_a (f(x)) \bigg] = \frac{1}{f(x). \ln(a)} \cdot f'(x)}$

Where:

• $a$ is the base of the logarithm,
• $f(x)$ is the argument of the logarithmic function, and
• $f'(x)$ is the derivative of $f(x)$.

For the natural logarithm (i.e., $a = e$), the formula simplifies to:

$\boxed{\frac{d}{dx} \bigg[ \ln\big(f(x)\big) \bigg] = \frac{f'(x)}{f(x)}}$

Example 1: Differentiating a Logarithmic Function with a Polynomial Argument

Differentiate $f(x) = \ln(3x^2 + 5x)$.

Solution:

Using the derivative of natural logarithms:

$f'(x) = \dfrac{1}{3x^2 + 5x} \cdot \dfrac{d}{dx}(3x^2 + 5x)$

The derivative of $3x^2 + 5x$ is $6x + 5$, so:

$f'(x) = \dfrac{6x + 5}{3x^2 + 5x}$

Thus, the final answer is: $\boxed{\dfrac{6x + 5}{3x^2 + 5x}}$.

Example 2: Differentiating a Logarithmic Function with an Exponential Argument

Differentiate $f(x) = \ln(e^{2x} + 1)$.

Solution:

Using the derivative of natural logarithms:

$f'(x) = \dfrac{1}{e^{2x} + 1} \cdot \dfrac{d}{dx}(e^{2x} + 1)$

The derivative of $e^{2x} + 1$ is $2e^{2x}$, so:

$f'(x) = \dfrac{2e^{2x}}{e^{2x} + 1}$

Thus, the final answer is: $\boxed{\dfrac{2e^{2x}}{e^{2x} + 1}}$.

Example 3: Differentiating a Logarithmic Function with a Power of $x$

Differentiate $f(x) = \log_2(x^3 + 4x)$.

Solution:

Using the derivative of logarithms with arbitrary base:

$f'(x) = \dfrac{1}{(x^3 + 4x) .\ln(2)} \cdot \dfrac{d}{dx}\big(x^3 + 4x\big)$

The derivative of $x^3 + 4x$ is $3x^2 + 4$, so:

$f'(x) = \dfrac{3x^2 + 4}{(x^3 + 4x) \ln(2)}$

Thus, the final answer is: $\boxed{\dfrac{3x^2 + 4}{(x^3 + 4x) .\ln(2)}}$.

Example 4: Finding the Rate of Change of a Logarithmic Function

Find the rate of change of the function $f(t) = \ln(5t^2 + 3t)$ at $t = 1$.

Solution:

Using the derivative of natural logarithms:

$f'(t) = \dfrac{1}{5t^2 + 3t} \cdot \dfrac{d}{dt}(5t^2 + 3t)$

The derivative of $5t^2 + 3t$ is $10t + 3$, so:

$f'(t) = \dfrac{10t + 3}{5t^2 + 3t}$

Now, substitute $t = 1$:

$f'(1) = \dfrac{10(1) + 3}{5(1)^2 + 3(1)} = \dfrac{10 + 3}{5 + 3} = \dfrac{13}{8}$

Thus, the rate of change of the function at $t = 1$ is: $\boxed{\frac{13}{8}}$.