Last updated on August 5th, 2025
We use the derivative of 9ln(x), which is 9/x, to understand how the logarithmic function changes with respect to x. Derivatives are crucial in calculating profit or loss in real-world scenarios. We will now discuss the derivative of 9ln(x) in detail.
The derivative of 9ln(x) is denoted as d/dx (9ln(x)) or (9ln(x))', and its value is 9/x. This derivative indicates how the function 9ln(x) changes in response to variations in x, highlighting its differentiability within its domain.
Key concepts include:
Logarithmic Function: ln(x) is the natural logarithm of x.
Constant Multiple Rule: Used for differentiating a constant multiplied by a function.
Reciprocal Function: The derivative of ln(x) is 1/x.
The derivative of 9ln(x) is expressed as d/dx (9ln(x)) or (9ln(x))'. The formula used to differentiate 9ln(x) is: d/dx (9ln(x)) = 9/x
This formula is applicable for all x > 0, as ln(x) is only defined for positive x.
We can derive the derivative of 9ln(x) using proofs. To demonstrate this, we will apply differentiation rules and logarithmic identities. Several methods can be employed, such as:
We will now show that differentiating 9ln(x) results in 9/x using the methods mentioned:
The derivative of 9ln(x) can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient.
To find the derivative of 9ln(x) using the first principle, consider f(x) = 9ln(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) = 9ln(x), we write f(x + h) = 9ln(x + h).
Substituting these into equation (1), f'(x) = limₕ→₀ [9ln(x + h) - 9ln(x)] / h = 9limₕ→₀ [ln(x + h) - ln(x)] / h = 9limₕ→₀ [ln((x + h)/x)] / h = 9limₕ→₀ ln[(1 + h/x)] / h
Using the limit property, limₕ→₀ ln(1 + u)/u = 1, f'(x) = 9 * 1/x Hence, f'(x) = 9/x, proved.
To prove the differentiation of 9ln(x) using the constant multiple rule, Consider the constant 9 and the function ln(x). Using the rule: d/dx [c·f(x)] = c·f'(x)
Here, c = 9 and f(x) = ln(x), f'(x) = d/dx [ln(x)] = 1/x Thus, d/dx [9ln(x)] = 9 * 1/x = 9/x.
The chain rule can also be used to differentiate 9ln(x). Consider y = 9ln(x) Let u = ln(x), then y = 9u
By chain rule: dy/dx = dy/du * du/dx dy/du = 9 and du/dx = 1/x
Therefore, dy/dx = 9 * 1/x = 9/x.
When a function is differentiated multiple times, the resulting derivatives are higher-order derivatives. Higher-order derivatives can be complex, similar to understanding a car's speed (first derivative) and its acceleration (second derivative). Higher-order derivatives offer insights into functions like 9ln(x).
For the first derivative, we write f′(x), indicating the slope or rate of change at a point. The second derivative, f′′(x), is derived from the first derivative. The third derivative, f′′′(x), follows from the second derivative, continuing this pattern.
For the nth derivative of 9ln(x), we denote it as fⁿ(x), providing insights into the rate of change for each derivative order.
When x is 0, the derivative is undefined as ln(x) is not defined for non-positive x. When x is 1, the derivative of 9ln(x) = 9/1 = 9.
Mistakes often occur when differentiating 9ln(x). These can be avoided by understanding the correct procedures. Here are some common mistakes and solutions:
Calculate the derivative of (9ln(x) · x²)
Here, we have f(x) = 9ln(x) · x².
Using the product rule, f'(x) = u′v + uv′ In the given equation, u = 9ln(x) and v = x².
Differentiate each term, u′= d/dx (9ln(x)) = 9/x v′= d/dx (x²) = 2x
Substituting into the given equation, f'(x) = (9/x) · x² + 9ln(x) · 2x
Simplify terms to get the final answer, f'(x) = 9x + 18xln(x)
Thus, the derivative of the specified function is 9x + 18xln(x).
We find the derivative of the given function by dividing it into two parts. First, we find the derivative of each part and then combine them using the product rule to get the final result.
A construction company uses the function y = 9ln(x) to model the pressure in a pipe at a distance x meters from the source. If x = 2 meters, measure the rate of change of pressure.
We have y = 9ln(x) (pressure in the pipe)...(1)
Differentiate equation (1) dy/dx = 9/x
Given x = 2, substitute this into the derivative dy/dx = 9/2
Hence, the rate of change of pressure at x = 2 meters is 4.5.
We calculate the rate of change at x = 2 meters as 4.5, indicating how the pressure changes with respect to distance from the source.
Derive the second derivative of the function y = 9ln(x).
First, find the first derivative, dy/dx = 9/x...(1)
Differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [9/x] = -9/x²
Therefore, the second derivative of the function y = 9ln(x) is -9/x².
We use a step-by-step process, starting with the first derivative. We then differentiate 9/x to obtain the second derivative, simplifying to find the final answer.
Prove: d/dx (ln²(x)) = 2ln(x)/x.
Start using the chain rule: Consider y = ln²(x) = [ln(x)]²
Differentiate using the chain rule: dy/dx = 2ln(x) · d/dx [ln(x)]
Since the derivative of ln(x) is 1/x, dy/dx = 2ln(x)/x Hence proved.
In this step-by-step process, we use the chain rule to differentiate the equation. We replace ln(x) with its derivative and simplify to derive the equation.
Solve: d/dx (ln(x)/x)
To differentiate the function, use the quotient rule: d/dx (ln(x)/x) = (d/dx (ln(x)) · x - ln(x) · d/dx(x))/x²
Substitute d/dx (ln(x)) = 1/x and d/dx (x) = 1 (1/x · x - ln(x) · 1)/x² = (1 - ln(x))/x²
Therefore, d/dx (ln(x)/x) = (1 - ln(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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