Last updated on July 16th, 2025
We use the derivative of 1/e^x, which is -1/e^x, as a measuring tool for how the function 1/e^x 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 1/e^x in detail.
We now understand the derivative of 1/e^x. It is commonly represented as d/dx (1/e^x) or (1/e^x)', and its value is -1/e^x. The function 1/e^x has a clearly defined derivative, indicating it is differentiable within its domain.
The key concepts are mentioned below:
Exponential Function: (e^x) is the base of the natural logarithm.
Negative Exponent Rule: Used to simplify 1/e^x as e^-x.
Chain Rule: Rule for differentiating composite functions like e^-x.
The derivative of 1/e^x can be denoted as d/dx (1/e^x) or (1/e^x)'.
The formula we use to differentiate 1/e^x is: d/dx (1/e^x) = -1/e^x (or) (1/e^x)' = -1/e^x The formula applies to all x as e^x is always positive.
We can derive the derivative of 1/e^x using proofs. To show this, we will use the rules of differentiation and exponent rules.
There are several methods we use to prove this, such as:
By First Principle Using Chain Rule We will now demonstrate that the differentiation of 1/e^x results in -1/e^x using the above-mentioned methods:
By First Principle The derivative of 1/e^x can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient.
To find the derivative of 1/e^x using the first principle, consider f(x) = 1/e^x. Its derivative can be expressed as the following limit.
f'(x) = limₕ→₀ [f(x + h) - f(x)] / h Given that f(x) = 1/e^x, we write f(x + h) = 1/e^(x + h).
Substituting these into the equation, f'(x) = limₕ→₀ [1/e^(x + h) - 1/e^x] / h = limₕ→₀ [(e^x - e^(x + h)) / (e^(x + h) e^x)] / h = limₕ→₀ [-e^x(e^h - 1) / (e^(x + h) h)] = -1/e^x
Using Chain Rule To prove the differentiation of 1/e^x using the chain rule, We use the formula:
1/e^x = e^-x
The derivative of e^-x is found using the chain rule: d/dx (e^-x) = -e^-x
Therefore, d/dx (1/e^x) = -1/e^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 1/e^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 1/e^x, we generally use f^n(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).
The function 1/e^x has no undefined points as e^x is never zero. When x is 0, the derivative of 1/e^x = -1/e^0, which is -1.
Students frequently make mistakes when differentiating 1/e^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 (1/e^x · e^x)
Here, we have f(x) = (1/e^x) · e^x.
Since 1/e^x · e^x = 1, the derivative is straightforward. f'(x) = d/dx (1) = 0.
Thus, the derivative of the specified function is 0.
We find the derivative of the given function by realizing that 1/e^x · e^x simplifies to 1. The derivative of a constant, such as 1, is always 0.
A chemical reaction decreases in concentration following the function C(x) = 1/e^x. Find the rate of change of concentration at x = 2.
We have C(x) = 1/e^x.
Now, we will differentiate the equation: dC/dx = -1/e^x. Given x = 2, dC/dx = -1/e^2.
Therefore, the rate of change of concentration at x = 2 is -1/e^2.
We find the rate of change of concentration at x = 2, which indicates how the concentration decreases at that point.
Derive the second derivative of the function y = 1/e^x.
The first step is to find the first derivative, dy/dx = -1/e^x.
Now we will differentiate again to get the second derivative: d²y/dx² = d/dx [-1/e^x] = 1/e^x.
Therefore, the second derivative of the function y = 1/e^x is 1/e^x.
We use the step-by-step process, where we start with the first derivative. Differentiating again, we find the second derivative.
Prove: d/dx (1/e^(2x)) = -2/e^(2x).
Let’s start using the chain rule: Consider y = 1/e^(2x) = e^(-2x).
To differentiate, we use the chain rule, dy/dx = -2e^(-2x).
Substituting y = 1/e^(2x), d/dx (1/e^(2x)) = -2/e^(2x).
Hence proved.
In this step-by-step process, we used the chain rule to differentiate the equation by considering the negative exponent. As a final step, we show how the result aligns with the original function.
Solve: d/dx (1/e^x - x)
To differentiate the function, we differentiate each term separately: d/dx (1/e^x - x) = d/dx (1/e^x) - d/dx (x) = -1/e^x - 1.
Therefore, d/dx (1/e^x - x) = -1/e^x - 1.
In this process, we differentiate each term in the function separately to find the final derivative.
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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