Last updated on August 5th, 2025
We use the derivative of e^-5x to understand how the function changes in response to a slight change in x. Derivatives are crucial in various real-life scenarios, such as calculating profit or loss. We will now discuss the derivative of e^-5x in detail.
We now understand the derivative of e^-5x. It is commonly represented as d/dx (e^-5x) or (e^-5x)', and its value is -5e^-5x.
The function e^-5x has a clearly defined derivative, indicating it is differentiable everywhere. The key concepts are mentioned below:
Exponential Function: e^-5x is an exponential function with a base e and a constant exponent.
Chain Rule: A rule for differentiating compositions of functions, which is often used when differentiating exponential functions.
Constant Multiplier Rule: A rule used to differentiate functions multiplied by a constant.
The derivative of e^-5x can be denoted as d/dx (e^-5x) or (e^-5x)'. The formula we use to differentiate e^-5x is: d/dx (e^-5x) = -5e^-5x The formula applies to all x.
We can derive the derivative of e^-5x using several methods.
To show this, we will use the chain rule and the properties of exponential functions.
Here are the methods we use to prove this: Using Chain Rule To prove the differentiation of e^-5x using the chain rule, Consider u(x) = -5x and the outer function v(u) = e^u.
The derivative of v with respect to u is v'(u) = e^u, and the derivative of u with respect to x is u'(x) = -5. Using the chain rule: (v(u(x)))' = v'(u(x)) * u'(x), d/dx (e^-5x) = e^-5x * (-5) = -5e^-5x.
Hence, proved.
Using Constant Multiplier Rule We will now prove the derivative of e^-5x using the constant multiplier rule.
Given that d/dx (e^x) = e^x, For y = e^-5x, let u = -5x, then dy/du = e^u and du/dx = -5.
By the chain rule, dy/dx = dy/du * du/dx = e^u * (-5) = -5e^-5x. Thus, the derivative of e^-5x is -5e^-5x.
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 e^-5x. 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 e^-5x, the pattern can be observed as fⁿ(x) = (-5)ⁿe^-5x, indicating the change in the rate of change.
When x approaches infinity, e^-5x approaches zero, meaning the derivative -5e^-5x also approaches zero. When x is 0, the derivative of e^-5x = -5e^0, which is -5.
Students frequently make mistakes when differentiating e^-5x. 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 e^-5x · sin(x).
Here, we have f(x) = e^-5x · sin(x). Using the product rule, f'(x) = u′v + uv′ In the given equation, u = e^-5x and v = sin(x).
Let’s differentiate each term, u′ = d/dx (e^-5x) = -5e^-5x v′ = d/dx (sin(x)) = cos(x) Substituting into the given equation, f'(x) = (-5e^-5x) · sin(x) + (e^-5x) · cos(x)
Let’s simplify terms to get the final answer, f'(x) = -5e^-5x sin(x) + e^-5x cos(x)
Thus, the derivative of the specified function is -5e^-5x sin(x) + e^-5x cos(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 uses the function y = e^-5x to model the decay of a chemical substance. Determine the rate of decay when x = 1.
We have y = e^-5x (rate of decay)...(1) Now, we will differentiate the equation (1) Take the derivative of e^-5x: dy/dx = -5e^-5x
Given x = 1 (substitute this into the derivative) dy/dx = -5e^-5(1) = -5e^-5 Hence, the rate of decay of the chemical substance when x = 1 is -5e^-5.
We find the rate of decay at x = 1 by substituting into the derivative. This indicates how quickly the substance is decaying at that specific point.
Derive the second derivative of the function y = e^-5x.
The first step is to find the first derivative, dy/dx = -5e^-5x...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-5e^-5x] = -5 d/dx [e^-5x] = -5(-5e^-5x) = 25e^-5x Therefore, the second derivative of the function y = e^-5x is 25e^-5x.
We use the step-by-step process, where we start with the first derivative.
We then differentiate the first derivative to find the second derivative.
Prove: d/dx (e^-5x)^2 = -10e^-10x.
Let’s start using the chain rule: Consider y = (e^-5x)^2 [e^-5x]^2 To differentiate, we use the chain rule: dy/dx = 2e^-5x · d/dx [e^-5x]
Since the derivative of e^-5x is -5e^-5x, dy/dx = 2e^-5x · (-5e^-5x) = -10e^-10x Hence proved.
In this step-by-step process, we used the chain rule to differentiate the equation.
Then, we replace e^-5x with its derivative. As a final step, we substitute back to derive the equation.
Solve: d/dx (e^-5x/x).
To differentiate the function, we use the quotient rule: d/dx (e^-5x/x) = (d/dx (e^-5x) · x - e^-5x · d/dx(x))/x²
We will substitute d/dx (e^-5x) = -5e^-5x and d/dx(x) = 1 (-5e^-5x · x - e^-5x · 1) / x² = (-5xe^-5x - e^-5x) / x² = -e^-5x(5x + 1) / x² Therefore, d/dx (e^-5x/x) = -e^-5x(5x + 1) / 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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