Last updated on July 21st, 2025
We use the derivative of 6ln(x), which is 6/x, as a measuring tool for how the logarithmic function changes in response to a slight change in x. Derivatives help us calculate profit or loss in real-life situations. We will now talk about the derivative of 6ln(x) in detail.
We now understand the derivative of 6ln x. It is commonly represented as d/dx (6ln x) or (6ln x)', and its value is 6/x. The function 6ln x has a clearly defined derivative, indicating it is differentiable within its domain.
The key concepts are mentioned below: Logarithmic Function: (ln(x) is the natural logarithm of x).
Constant Multiple Rule: When differentiating a constant multiplied by a function.
Derivative of ln(x): The derivative of ln(x) is 1/x.
The derivative of 6ln x can be denoted as d/dx (6ln x) or (6ln x)'. The formula we use to differentiate 6ln x is: d/dx (6ln x) = 6/x The formula applies to all x where x > 0.
We can derive the derivative of 6ln x using proofs. To show this, we will use the rules of differentiation.
There are several methods we use to prove this, such as:
We will now demonstrate that the differentiation of 6ln x results in 6/x using the above-mentioned methods:
The derivative of 6ln x can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient.
To find the derivative of 6ln x using the first principle, we will consider f(x) = 6ln x. Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1)
Given that f(x) = 6ln x, we write f(x + h) = 6ln (x + h).
Substituting these into equation (1), f'(x) = limₕ→₀ [6ln(x + h) - 6ln x] / h = limₕ→₀ [6(ln(x + h) - ln x)] / h = 6 · limₕ→₀ [ln(x + h) - ln x] / h
Using the property of logarithms: ln a - ln b = ln(a/b), = 6 · limₕ→₀ ln((x + h)/x) / h = 6 · limₕ→₀ ln(1 + h/x) / h
As h approaches 0, ln(1 + h/x) ≈ h/x, f'(x) = 6 · (1/x) = 6/x
Hence, proved.
To prove the differentiation of 6ln x using the constant multiple rule, We use the formula: 6ln x = 6 · ln x The derivative of ln x is 1/x. So, the derivative of 6ln x is: d/dx (6ln x) = 6 · d/dx (ln x) = 6 · (1/x) = 6/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 6ln(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 6ln(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. (continuing for higher-order derivatives).
When x is 0, the derivative is undefined because ln(x) is undefined there. When x is 1, the derivative of 6ln x = 6/1, which is 6.
Students frequently make mistakes when differentiating 6ln x. These mistakes can be resolved by understanding the proper solutions. Here are a few common mistakes and ways to solve them:
Calculate the derivative of (6ln x·x²)
Here, we have f(x) = 6ln x·x². Using the product rule, f'(x) = u′v + uv′ In the given equation, u = 6ln x and v = x². Let’s differentiate each term, u′= d/dx (6ln x) = 6/x v′= d/dx (x²) = 2x
substituting into the given equation, f'(x) = (6/x)·x² + (6ln x)·2x
Let’s simplify terms to get the final answer, f'(x) = 6x + 12xln x
Thus, the derivative of the specified function is 6x + 12xln x.
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.
A company tracks the growth of bacteria using the function y = 6ln(x) where y represents the growth rate at time x. If x = 2 hours, measure the growth rate.
We have y = 6ln(x) (growth rate)...(1)
Now, we will differentiate the equation (1) Take the derivative 6ln(x): dy/dx = 6/x Given x = 2 (substitute this into the derivative) dy/dx = 6/2 = 3
Hence, we get the growth rate at time x = 2 hours as 3.
We find the growth rate at x = 2 hours as 3, which means that at this point, the growth rate is 3 units per hour.
Derive the second derivative of the function y = 6ln(x).
The first step is to find the first derivative, dy/dx = 6/x...(1)
Now, we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [6/x] = -6/x²
Therefore, the second derivative of the function y = 6ln(x) is -6/x².
We use the step-by-step process, where we start with the first derivative. Then, we differentiate 6/x directly to obtain the second derivative.
Prove: d/dx ((6ln x)²) = 12ln(x)/x.
Let’s start using the chain rule: Consider y = (6ln x)²
To differentiate, we use the chain rule: dy/dx = 2(6ln x)·d/dx (6ln x)
Since the derivative of 6ln x is 6/x, dy/dx = 2(6ln x)·(6/x) = 72ln(x)/x
Hence proved.
In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replace 6ln x with its derivative. As a final step, we simplify to derive the equation.
Solve: d/dx (6ln x/x)
To differentiate the function, we use the quotient rule: d/dx (6ln x/x) = (d/dx (6ln x)·x - 6ln x·d/dx(x))/x²
We will substitute d/dx (6ln x) = 6/x and d/dx (x) = 1 = (6x/x - 6ln x·1)/x² = (6 - 6ln x)/x²
Therefore, d/dx (6ln x/x) = (6 - 6ln x)/x²
In this process, we differentiate the given function using the quotient rule. As a final step, we simplify the equation to obtain the final result.
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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