Last updated on July 15th, 2025
We use the derivative of 5ln(x), which is 5/x, as a tool for understanding how the logarithmic function changes in response to changes in x. Derivatives are crucial in various fields, including economics and engineering, to calculate rates of change. Let's explore the derivative of 5ln(x) in detail.
The derivative of 5ln(x) is commonly represented as d/dx (5lnx) or (5lnx)', and its value is 5/x. The function 5ln(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 Multiplier Rule: Rule for differentiating functions multiplied by a constant. Derivative of ln(x): The basic derivative of ln(x) is 1/x.
The derivative of 5ln(x) can be denoted as d/dx (5lnx) or (5lnx)'. The formula we use to differentiate 5ln(x) is: d/dx (5lnx) = 5 * d/dx (lnx) = 5/x The formula applies to all x>0.
We can derive the derivative of 5ln(x) using proofs. To show this, we will use the rules of differentiation. There are several methods we use to prove this, such as: By First Principle Using the Constant Multiplier Rule By First Principle The derivative of 5ln(x) can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient. To find the derivative of 5ln(x) using the first principle, we consider f(x) = 5ln(x). Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h Given that f(x) = 5ln(x), we write f(x + h) = 5ln(x + h). Substituting these into the equation, f'(x) = limₕ→₀ [5ln(x + h) - 5ln(x)] / h = 5 * limₕ→₀ [ln(x + h) - ln(x)] / h = 5 * limₕ→₀ [ln((x + h)/x)] / h = 5 * limₕ→₀ [ln(1 + h/x)] / h = 5 * (1/x) * limₕ→₀ [ln(1 + u)] / u (where u = h/x) Using limit formulas, limₕ→₀ ln(1 + u)/u = 1. f'(x) = 5 * (1/x) = 5/x Hence, proved. Using the Constant Multiplier Rule To prove the differentiation of 5ln(x) using the constant multiplier rule, We use the formula: d/dx (c * f(x)) = c * d/dx (f(x)) Let f(x) = ln(x) and c = 5. The derivative of ln(x) is: d/dx (ln(x)) = 1/x So, d/dx (5ln(x)) = 5 * (1/x) = 5/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's speed (first derivative) and the rate at which the speed changes (second derivative). Higher-order derivatives make it easier to understand functions like 5ln(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 5ln(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 approaches 0, the derivative becomes undefined because ln(x) is undefined for non-positive values. When x = 1, the derivative of 5ln(x) = 5/1, which is 5.
Students frequently make mistakes when differentiating 5ln(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 (5lnx * x^3).
Here, we have f(x) = 5lnx * x^3. Using the product rule, f'(x) = u′v + uv′ In the given equation, u = 5lnx and v = x^3. Let’s differentiate each term, u′= d/dx (5lnx) = 5/x v′= d/dx (x^3) = 3x^2 Substituting into the given equation, f'(x) = (5/x) * x^3 + 5lnx * 3x^2 Let’s simplify terms to get the final answer, f'(x) = 5x^2 + 15x^2lnx Thus, the derivative of the specified function is 5x^2 + 15x^2lnx.
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 population of bacteria grows exponentially, and its growth rate is modeled by P(t) = 5ln(t) where P represents the population at time t. If t = 2 hours, find the growth rate.
We have P(t) = 5ln(t) (rate of growth)...(1) Now, we will differentiate the equation (1) Take the derivative of 5ln(t): dP/dt = 5/t Given t = 2 (substitute this into the derivative) dP/dt = 5/2 Hence, we get the growth rate of the population at t = 2 hours as 2.5.
We find the growth rate at t = 2 hours as 2.5, which indicates the rate of population increase at that point in time.
Derive the second derivative of the function y = 5ln(x).
The first step is to find the first derivative, dy/dx = 5/x...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [5/x] d²y/dx² = -5/x² Therefore, the second derivative of the function y = 5ln(x) is -5/x².
We use the step-by-step process, where we start with the first derivative. We then differentiate again to find the second derivative, simplifying to get the final answer.
Prove: d/dx ((5lnx)²) = 10ln(x)/x.
Let’s start using the chain rule: Consider y = (5lnx)² To differentiate, we use the chain rule: dy/dx = 2 * (5lnx) * d/dx (5lnx) Since the derivative of 5ln(x) is 5/x, dy/dx = 2 * (5lnx) * (5/x) = 10ln(x)/x Hence proved.
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 and simplify to derive the equation.
Solve: d/dx (5lnx/x).
To differentiate the function, we use the quotient rule: d/dx (5lnx/x) = (d/dx (5lnx) * x - 5lnx * d/dx(x))/x² We will substitute d/dx (5lnx) = 5/x and d/dx(x) = 1 = (5/x * x - 5lnx * 1) / x² = (5 - 5lnx) / x² Therefore, d/dx (5lnx/x) = (5 - 5lnx) / x²
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.
Derivative: The derivative of a function indicates how the given function changes in response to a slight change in x. Logarithmic Function: A function that uses the natural logarithm, represented as ln(x). Constant Multiplier Rule: A rule used in differentiation where a constant multiplied by a function results in the constant multiplied by the derivative of the function. First Derivative: The initial result of differentiating a function, which gives the rate of change of a specific function. Quotient Rule: A rule used to differentiate functions that are divided by each other, expressed as d/dx (u/v) = (v * u' - u * v') / v².
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