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Last updated on September 9, 2025
We use the derivative of log(e), which is 1, as a tool to understand how the natural logarithmic function changes in response to a slight change in x. Derivatives have practical applications in fields like finance or physics. We will now discuss the derivative of log(e) in detail.
We now understand the derivative of log e.
It is commonly represented as d/dx (log(e)) or (log(e))', and its value is 1.
The function log(e) has a clearly defined derivative, indicating it is differentiable within its domain.
The key concepts are mentioned below:
Logarithmic Function: log(e) is the natural logarithm of e, which equals 1.
Constant Function: A function that returns the same value, which in this case is 1, regardless of x.
Derivative of a Constant: The derivative of any constant function is 0.
The derivative of log e can be denoted as d/dx (log e) or (log e)'. The formula we use to differentiate log e is: d/dx (log e) = 0 The formula applies because log e is a constant value.
We can derive the derivative of log e using proofs.
To show this, we consider that log e is a constant value.
There are several methods we use to prove this, such as:
By Definition of a Constant
Using Fundamental Theorem of Calculus
Using Properties of Logarithms
We will now demonstrate that the differentiation of log e results in 0 using the above-mentioned methods:
By Definition of a Constant The derivative of log e can be proved using the definition of the derivative of a constant, which is zero.
To find the derivative of log e, we consider f(x) = log e.
Its derivative can be expressed as: f'(x) = d/dx (log e) = 0
This is because the derivative of any constant value is 0.
Using Fundamental Theorem of Calculus
To prove the differentiation of log e using the Fundamental Theorem of Calculus, We note that log e = 1, a constant.
Therefore, by the theorem: d/dx (c) = 0 for any constant c.
Hence, the derivative of log e is 0.
Using Properties of Logarithms
We will now prove the derivative of log e using properties of logarithms.
Here, we use the formula: log e = 1 Given that, the derivative of a constant is 0, we have: d/dx (1) = 0.
Thus, the derivative of log e is 0.
When a function is differentiated several times, the derivatives obtained are referred to as higher-order derivatives.
Higher-order derivatives for a constant function like log e are straightforward.
Since the first derivative is already 0, all subsequent derivatives will also be 0.
This means that no matter how many times you differentiate log e, the result will always be 0.
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 log(e), we generally use fⁿ(x) for the nth derivative of a function f(x), and it remains 0.
Since log e is a constant, there are no special cases where the derivative differs. The derivative remains 0 for all x.
Students frequently make mistakes when differentiating log e. 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 (log e · x²)
Here, we have f(x) = log e · x².
Since log e is a constant, its derivative is 0.
Using the product rule: f'(x) = (d/dx (log e)) · x² + log e · d/dx (x²) = 0 · x² + 1 · 2x = 2x
Thus, the derivative of the specified function is 2x.
We find the derivative of the given function by recognizing that log e is a constant. We then use the product rule and differentiate x² to get the final result.
A company is modeling its constant growth using the function y = log(e) + 5x. If x = 10, find the rate of change of the function.
We have y = log(e) + 5x. Now, we will differentiate the equation:
dy/dx = d/dx (log(e)) + d/dx (5x) = 0 + 5
The rate of change of the function is 5, regardless of x.
We find the rate of change by differentiating the function. Since log e is a constant, its derivative is 0, and we differentiate 5x to find the rate of change.
Derive the second derivative of the function y = log(e) + 3x².
The first step is to find the first derivative:
dy/dx = d/dx (log(e)) + d/dx (3x²) = 0 + 6x
Now we will differentiate again to get the second derivative: d²y/dx² = d/dx (6x) = 6
Therefore, the second derivative of the function y = log(e) + 3x² is 6.
We use the step-by-step process, starting with the first derivative. Since log e is constant, we differentiate the remaining part. Then, we differentiate again to find the second derivative.
Prove: d/dx (log(e) · tan(x)) = tan(x) · 0 + log(e) · sec²(x).
Let’s start using the product rule:
Consider y = log(e) · tan(x)
To differentiate, we use the product rule:
dy/dx = d/dx (log(e)) · tan(x) + log(e) · d/dx (tan(x))
Since the derivative of log(e) is 0 and d/dx(tan(x)) is sec²(x), dy/dx = 0 · tan(x) + 1 · sec²(x) dy/dx = sec²(x)
Hence proved.
In this step-by-step process, we used the product rule to differentiate the equation. We replace log(e) with its derivative and differentiate tan(x) to find the result.
Solve: d/dx (log(e)/x)
To differentiate the function, we use the quotient rule:
d/dx (log(e)/x) = (d/dx (log(e)) · x - log(e) · d/dx(x)) / x²
We substitute d/dx (log(e)) = 0 and d/dx(x) = 1: = (0 · x - 1 · 1) / x² = -1/x²
Therefore, d/dx (log(e)/x) = -1/x²
In this process, we differentiate the given function using the quotient rule. We recognize log(e) as a constant and 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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