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107 LearnersLast updated on 26 September 2025
Derivative of ln(f(x))
We use the derivative of ln(f(x)), which is f'(x)/f(x), as a tool for understanding how the natural logarithm function changes in response to a slight change in x. Derivatives help us calculate profit or loss in real-life situations. We will now discuss the derivative of ln(f(x)) in detail.
What is the Derivative of ln(f(x))?
We now understand the derivative of ln(f(x)). It is commonly represented as d/dx (ln(f(x))) or (ln(f(x)))', and its value is f'(x)/f(x). The function ln(f(x)) has a clearly defined derivative, indicating it is differentiable within its domain.
The key concepts are mentioned below: Natural Logarithm Function: ln(x) is the logarithm to the base e. Chain Rule: A rule for differentiating composite functions like ln(f(x)), which requires the derivative of the inner function. Quotient Function: Involves the derivative being expressed as a quotient of functions.
Derivative of ln(f(x)) Formula
The derivative of ln(f(x)) can be denoted as d/dx (ln(f(x))) or (ln(f(x)))'.
The formula we use to differentiate ln(f(x)) is: d/dx (ln(f(x))) = f'(x)/f(x) The formula applies to all x for which f(x) > 0.
Proofs of the Derivative of ln(f(x))
We can derive the derivative of ln(f(x)) using proofs. To show this, we will use the properties of logarithms along with the rules of differentiation. There are several methods we use to prove this, such as: By First Principle Using Chain Rule We will now demonstrate that the differentiation of ln(f(x)) results in f'(x)/f(x) using the above-mentioned methods:
By First Principle The derivative of ln(f(x)) can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient. To find the derivative of ln(f(x)) using the first principle, we will consider g(x) = ln(f(x)). Its derivative can be expressed as the following limit. g'(x) = limₕ→₀ [g(x + h) - g(x)] / h … (1) Given that g(x) = ln(f(x)), we write g(x + h) = ln(f(x + h)). Substituting these into equation (1), g'(x) = limₕ→₀ [ln(f(x + h)) - ln(f(x))] / h = limₕ→₀ ln[(f(x + h)/f(x))] / h
Using the properties of logarithms, ln(a) - ln(b) = ln(a/b), g'(x) = limₕ→₀ ln[1 + (f(x + h) - f(x))/f(x)] / h Using the approximation ln(1 + u) ≈ u for small u, g'(x) = limₕ→₀ (f(x + h) - f(x))/(h f(x)) g'(x) = f'(x)/f(x) Hence, proved.
Using Chain Rule To prove the differentiation of ln(f(x)) using the chain rule, We use the formula: ln(f(x)) = ln(u), where u = f(x) The derivative of ln(u) is 1/u * du/dx Let u = f(x) Then, d/du(ln(u)) = 1/u And, du/dx = f'(x) Therefore, d/dx (ln(f(x))) = 1/f(x) * f'(x) = f'(x)/f(x)
Higher-Order Derivatives of ln(f(x))
When a function is differentiated several times, the derivatives obtained are referred to as higher-order derivatives. Higher-order derivatives can be a little tricky. To understand them better, think of a car where the speed changes (first derivative) and the rate at which the speed changes (second derivative) also changes. Higher-order derivatives make it easier to understand functions like ln(f(x)).
For the first derivative of a function, we write g′(x), which indicates how the function changes or its slope at a certain point. The second derivative is derived from the first derivative, which is denoted using g′′(x). Similarly, the third derivative, g′′′(x), is the result of the second derivative, and this pattern continues.
For the nth Derivative of ln(f(x)), we generally use gⁿ(x) for the nth derivative of a function g(x), which tells us the change in the rate of change.
Special Cases:
When f(x) is a constant, the derivative is zero because the natural logarithm function becomes a constant. If f(x) is e^x, then the derivative of ln(e^x) = x is 1.
Common Mistakes and How to Avoid Them in Derivatives of ln(f(x))
Students frequently make mistakes when differentiating ln(f(x)). These mistakes can be resolved by understanding the proper solutions. Here are a few common mistakes and ways to solve them:
Mistake 1
Not applying the chain rule
Students may forget to apply the chain rule, which can lead to incorrect results. They often skip the derivative of the inner function, especially when dealing with composite functions.
Ensure that each step is written in order. Students might find it tedious, but it is important to avoid errors in the process.
Mistake 2
Ignoring the domain of f(x)
They might not remember that ln(f(x)) is undefined for f(x) ≤ 0. Keep in mind that you should consider the domain of the function that you differentiate.
It will help you understand that the function is not continuous at such points.
Mistake 3
Incorrect use of the quotient rule
While differentiating functions like ln(f(x))/x, students misapply the quotient rule.
For example: Incorrect differentiation: d/dx (ln(f(x))/x) = f'(x)/(f(x)x²). d/dx (u/v) = (v · u' - u · v')/v² (where u = ln(f(x)) and v = x) Applying the quotient rule, d/dx (ln(f(x))/x) = (x · f'(x)/f(x) - ln(f(x)))/x² To avoid this mistake, write the quotient rule without errors. Always check for errors in the calculation and ensure it is properly simplified.
Mistake 4
Not writing constants and coefficients
There is a common mistake that students at times forget to multiply the constants placed before ln(f(x)).
1qqqqqqqqqq211111qq11q1qq1qqFor example, they incorrectly write d/dx (5 ln(f(x))) = f'(x)/f(x). Students should check the constants in the terms and ensure they are multiplied properly. For example, the correct equation is d/dx (5 ln(f(x))) = 5 f'(x)/f(x).
Mistake 5
Not applying the product rule
```````Students often forget to use the product rule. This happens when the derivative of the product of functions is not considered.
For example: Incorrect: d/dx (ln(f(x)) · g(x)) = f'(x)/f(x) · g(x).
To fix this error, students should apply the product rule: d/dx (ln(f(x)) · g(x)) = ln(f(x))' · g(x) + ln(f(x)) · g'(x).
Examples Using the Derivative of ln(f(x))
Problem 1
Calculate the derivative of ln(x² + 1) · e^x
Here, we have f(x) = ln(x² + 1) · ex. Using the product rule, f'(x) = u′v + uv′ In the given equation, u = ln(x² + 1) and v = ex. Let’s differentiate each term, u′ = d/dx (ln(x² + 1)) = (2x)/(x² + 1) v′ = d/dx (ex) = ex
Substituting into the given equation, f'(x) = [(2x)/(x² + 1)] · ex + ln(x² + 1) · ex Let’s simplify terms to get the final answer, f'(x) = (2x · ex)/(x² + 1) + ln(x² + 1) · ex
Thus, the derivative of the specified function is (2x · ex)/(x² + 1) + ln(x² + 1) · ex.
Explanation
We find the derivative of the given function by dividing the function into two parts. The first step is finding its derivative and then combining them using the product rule to get the final result.
Problem 2
A car's velocity is represented by the function v(x) = ln(x + 1) where x represents time in seconds. If x = 3 seconds, measure the rate of change of velocity.
We have v(x) = ln(x + 1) (velocity of the car)...(1)Now, we will differentiate the equation (1) Take the derivative ln(x + 1): dv/dx = 1/(x + 1)Given x = 3 (substitute this into the derivative) dv/dx = 1/(3 + 1) = 1/4Hence, we get the rate of change of velocity at x = 3 seconds as 1/4.
Explanation
We find the rate of change of velocity at x = 3 seconds as 1/4, which means that at this point, the rate of change of velocity is 1/4 units per second.
Problem 3
Derive the second derivative of the function y = ln(x² + 1).
The first step is to find the first derivative, dy/dx = (2x)/(x² + 1)...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [(2x)/(x² + 1)] Here we use the quotient rule, d²y/dx² = [(x² + 1) · 2 - 2x · 2x]/(x² + 1)² = [2x² + 2 - 4x²]/(x² + 1)² = [-2x² + 2]/(x² + 1)² Therefore, the second derivative of the function y = ln(x² + 1) is [-2x² + 2]/(x² + 1)².
Explanation
We use the step-by-step process, where we start with the first derivative.
Using the quotient rule, we differentiate the first derivative.
We then simplify the terms to find the final answer.
Problem 4
Prove: d/dx (ln(x)²) = 2 ln(x)/x.
Let’s start using the chain rule: Consider y = ln(x)² = [ln(x)]² To differentiate, we use the chain rule: dy/dx = 2 ln(x) · d/dx [ln(x)] Since the derivative of ln(x) is 1/x, dy/dx = 2 ln(x) · (1/x) Substituting y = ln(x)², d/dx (ln(x)²) = 2 ln(x)/x Hence proved.
Explanation
In this step-by-step process, we used the chain rule to differentiate the equation.
Then, we replace ln(x) with its derivative.
As a final step, we substitute y = ln(x)² to derive the equation.
Problem 5
Solve: d/dx (ln(x + 1)/x)
To differentiate the function, we use the quotient rule: d/dx (ln(x + 1)/x) = (d/dx (ln(x + 1)) · x - ln(x + 1) · d/dx(x))/x² We will substitute d/dx (ln(x + 1)) = 1/(x + 1) and d/dx(x) = 1 = [(1/(x + 1)) · x - ln(x + 1) · 1]/x² = [x/(x + 1) - ln(x + 1)]/x² Therefore, d/dx (ln(x + 1)/x) = [x/(x + 1) - ln(x + 1)]/x²
Explanation
In this process, we differentiate the given function using the product rule and quotient rule.
As a final step, we simplify the equation to obtain the final result.
FAQs on the Derivative of ln(f(x))
1.Find the derivative of ln(f(x)).
2.Can we use the derivative of ln(f(x)) in real life?
3.Is it possible to take the derivative of ln(f(x)) at the point where f(x) = 0?
4.What rule is used to differentiate ln(f(x))/x?
5.Are the derivatives of ln(f(x)) and ln⁻¹(x) the same?
6.Can we find the derivative of the ln(f(x)) formula?
Important Glossaries for the Derivative of ln(f(x))
- Derivative: The derivative of a function indicates how the given function changes in response to a slight change in x.
- Natural Logarithm: A logarithm to the base e, denoted as ln(x).
- Chain Rule: A rule for finding the derivative of the composition of two or more functions.
- Quotient Rule: A rule for finding the derivative of a quotient of two functions.
- Higher-Order Derivatives: Derivatives of a function taken multiple times, such as the second derivative or third derivative.
Jaskaran Singh Saluja
About the Author
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
Fun Fact
: He loves to play the quiz with kids through algebra to make kids love it.
