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105 LearnersLast updated on October 8, 2025
Derivative of ln(3-x)
We use the derivative of ln(3-x), which is -1/(3-x), as a tool for understanding how the natural logarithmic function changes in response to a slight change in x. Derivatives are crucial for various applications in mathematics and real-life situations, such as calculating rates of change. We will now discuss the derivative of ln(3-x) in detail.
What is the Derivative of ln(3-x)?
We now understand the derivative of ln(3-x). It is commonly represented as d/dx [ln(3-x)] or [ln(3-x)]', and its value is -1/(3-x). The function ln(3-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 natural logarithm of x.
Chain Rule: Rule for differentiating composite functions, such as ln(3-x).
Derivative Formula: The formula for differentiating ln(3-x) involves the chain rule.
Derivative of ln(3-x) Formula
The derivative of ln(3-x) can be denoted as d/dx [ln(3-x)] or [ln(3-x)]'.
The formula used to differentiate ln(3-x) is: d/dx [ln(3-x)] = -1/(3-x)
The formula applies to all x where 3-x > 0, i.e., x < 3.
Proofs of the Derivative of ln(3-x)
We can derive the derivative of ln(3-x) using proofs. To show this, we will use the rules of differentiation, particularly the chain rule. Here are the methods to prove this:
Using Chain Rule
To prove the differentiation of ln(3-x) using the chain rule: Let u = 3-x. Therefore, ln(3-x) = ln(u). The derivative of ln(u) with respect to u is 1/u. We have du/dx = -1 (since the derivative of 3-x is -1). Using the chain rule: d/dx [ln(3-x)] = d/du [ln(u)] * du/dx = (1/u) * (-1) = -1/u. Substitute u = 3-x back into the equation: d/dx [ln(3-x)] = -1/(3-x). Hence, proved.
Higher-Order Derivatives of ln(3-x)
When a function is differentiated several times, the derivatives obtained are referred to as higher-order derivatives. 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(3-x).
For the first derivative of a function, we write f′(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 f′′(x). Similarly, the third derivative, f′′′(x), is the result of the second derivative and this pattern continues.
For the nth Derivative of ln(3-x), we generally use fⁿ(x) for the nth derivative of a function f(x) which tells us the change in the rate of change.
Special Cases:
When x approaches 3, the derivative is undefined because ln(3-x) becomes undefined there. When x is 0, the derivative of ln(3-x) is -1/3.
Common Mistakes and How to Avoid Them in Derivatives of ln(3-x)
Students frequently make mistakes when differentiating ln(3-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 simplifying the equation
Students may forget to simplify the equation, which can lead to incomplete or incorrect results. They often skip steps and directly arrive at the result, especially when using the chain rule.
Ensure that each step is written in order. Although it may seem tedious, it is important to avoid errors in the process.
Mistake 2
Forgetting the Domain Restrictions
They might not remember that ln(3-x) is undefined for x ≥ 3. 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 certain points.
Mistake 3
Incorrect use of Chain Rule
While differentiating functions such as ln(3-x), students misapply the chain rule. For example: Incorrect differentiation: d/dx [ln(3-x)] = 1/(3-x). Applying the chain rule correctly: d/dx [ln(3-x)] = (1/(3-x)) * (-1) = -1/(3-x).
To avoid this mistake, ensure the chain rule is applied correctly by identifying the inner function and correctly differentiating it.
Mistake 4
Not writing Constants and Coefficients
There is a common mistake that students at times forget to multiply the constants placed before the variable. For example, they incorrectly write d/dx [c ln(3-x)] as c * 1/(3-x).
Students should check the constants in the terms and ensure they are multiplied properly. For example, the correct equation is d/dx [c ln(3-x)] = c * (-1)/(3-x).
Mistake 5
Not Applying the Chain Rule
Students often forget to use the chain rule. This happens when the derivative of the inner function is not considered. For example: Incorrect: d/dx [ln(3-x²)] = 1/(3-x²).
To fix this error, students should divide the functions into inner and outer parts. Then, make sure that each function is differentiated. For example, d/dx [ln(3-x²)] = (1/(3-x²)) * (-2x).
Examples Using the Derivative of ln(3-x)
Problem 1
Calculate the derivative of ln(3-x)².
Here, we have f(x) = ln(3-x)². Using the chain rule, f'(x) = 2 * ln(3-x) * d/dx [ln(3-x)]. We already know that d/dx [ln(3-x)] = -1/(3-x). So, f'(x) = 2 * ln(3-x) * (-1/(3-x)) = -2 * ln(3-x)/(3-x). Thus, the derivative of the specified function is -2 * ln(3-x)/(3-x).
Explanation
We find the derivative of the given function by applying the chain rule.
The first step is finding its derivative and then combining them to get the final result.
Problem 2
An exponential growth model is represented by the function y = ln(3-x) where y represents the growth rate at a certain time x. If x = 1, calculate the growth rate.
We have y = ln(3-x) (growth rate)...(1) Now, we will differentiate equation (1) Take the derivative ln(3-x): dy/dx = -1/(3-x) Given x = 1, substitute this into the derivative: dy/dx = -1/(3-1) = -1/2. Hence, the growth rate at x=1 is -1/2.
Explanation
We find the growth rate at x=1 as -1/2, which means that at that time, the growth rate decreases by half per unit increase in x.
Problem 3
Derive the second derivative of the function y = ln(3-x).
The first step is to find the first derivative, dy/dx = -1/(3-x)...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-1/(3-x)] Here we use the chain rule, d²y/dx² = (-1) * d/dx [1/(3-x)] d²y/dx² = (-1) * (-1) * (1/(3-x)²) = 1/(3-x)². Therefore, the second derivative of the function y = ln(3-x) is 1/(3-x)².
Explanation
We use a step-by-step process, where we start with the first derivative.
Using the chain rule, we differentiate -1/(3-x).
We then substitute and simplify the terms to find the final answer.
Problem 4
Prove: d/dx [ln((3-x)²)] = -2/(3-x).
Let’s start using the chain rule: Consider y = ln((3-x)²). Using the chain rule: dy/dx = 1/((3-x)²) * d/dx [(3-x)²]. The derivative of (3-x)² is -2(3-x). So, dy/dx = 1/((3-x)²) * (-2(3-x)) = -2/(3-x). Hence proved.
Explanation
In this step-by-step process, we used the chain rule to differentiate the equation.
Then, we replace the inner function with its derivative and simplify to derive the equation.
Problem 5
Solve: d/dx [ln(3-x)/(x+1)]
To differentiate the function, we use the quotient rule: d/dx [ln(3-x)/(x+1)] = (d/dx [ln(3-x)] * (x+1) - ln(3-x) * d/dx [x+1])/(x+1)² We know d/dx [ln(3-x)] = -1/(3-x) and d/dx [x+1] = 1. = ([-1/(3-x)] * (x+1) - ln(3-x) * 1)/(x+1)² = [-(x+1)/(3-x) - ln(3-x)]/(x+1)² = [-(x+1) - (3-x)ln(3-x)]/((3-x)(x+1)²). Therefore, d/dx [ln(3-x)/(x+1)] = [-(x+1) - (3-x)ln(3-x)]/((3-x)(x+1)²).
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(3-x)
1.Find the derivative of ln(3-x).
2.Can we use the derivative of ln(3-x) in real life?
3.Is it possible to take the derivative of ln(3-x) at the point where x = 3?
4.What rule is used to differentiate ln(3-x)/(x+1)?
5.Are the derivatives of ln(3-x) and ln(x) the same?
Important Glossaries for the Derivative of ln(3-x)
- Derivative: The derivative of a function indicates how the given function changes in response to a slight change in x.
- Natural Logarithm: The natural logarithm is the logarithm to the base e, where e is an irrational and transcendental constant approximately equal to 2.718281828459.
- Chain Rule: A rule for differentiating compositions of functions, where the derivative of the composition is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.
- Quotient Rule: A formula for finding the derivative of a quotient of two functions.
- Domain: The set of input values (x-values) for which a function is defined.
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Jaskaran Singh Saluja
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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.
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