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412 LearnersLast updated on August 5, 2025

We now understand the derivative of e^-8x. It is commonly represented as d/dx (e^-8x) or (e^-8x)', and its value is -8e^-8x. The function e^-8x has a clearly defined derivative, indicating it is differentiable within its domain. The key concepts are mentioned below: Exponential Function: (e^x is an exponential function). Chain Rule: A rule for differentiating compositions of functions (used here for e^-8x). Constant Multiplication Rule: The derivative of a constant times a function is the constant times the derivative of the function.
The derivative of e^-8x can be denoted as d/dx (e^-8x) or (e^-8x)'. The formula we use to differentiate e^-8x is: d/dx (e^-8x) = -8e^-8x The formula applies to all x.
We can derive the derivative of e^-8x using proofs. To show this, we will use the rules of differentiation. There are several methods we use to prove this, such as: Using Chain Rule Using Constant Multiplication Rule We will now demonstrate that the differentiation of e^-8x results in -8e^-8x using the above-mentioned methods: Using Chain Rule To prove the differentiation of e^-8x using the chain rule, We use the formula: Let u = -8x, then the function becomes e^u. The derivative of e^u with respect to u is e^u. The derivative of u = -8x with respect to x is -8. By chain rule: d/dx (e^u) = e^u * (du/dx) d/dx (e^-8x) = e^-8x * (-8) d/dx (e^-8x) = -8e^-8x Hence, proved. Using Constant Multiplication Rule We will now prove the derivative of e^-8x using the constant multiplication rule. Given that, d/dx (e^-8x) = d/dx (e^(-8x)) = -8 * d/dx (e^x) = -8 * e^x Substituting back, d/dx (e^-8x) = -8e^-8x Thus, the derivative of e^-8x is -8e^-8x.


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^-8x. 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^-8x, 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 infinity, the derivative approaches 0 because the exponential function decays. At x = 0, the derivative of e^-8x = -8e^0, which is -8.
Students frequently make mistakes when differentiating e^-8x. 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^-8x(2x + 3).
Here, we have f(x) = e^-8x(2x + 3). Using the product rule, f'(x) = u′v + uv′ In the given equation, u = e^-8x and v = 2x + 3. Let’s differentiate each term, u′ = d/dx (e^-8x) = -8e^-8x v′ = d/dx (2x + 3) = 2 Substituting into the given equation, f'(x) = (-8e^-8x)(2x + 3) + (e^-8x)(2) Let’s simplify terms to get the final answer, f'(x) = -8e^-8x(2x + 3) + 2e^-8x Thus, the derivative of the specified function is -8e^-8x(2x + 3) + 2e^-8x.
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 decreases exponentially, modeled by P(t) = e^-8t. Calculate the rate of change of the population when t = 2 hours.
We have P(t) = e^-8t (population model)...(1) Now, we will differentiate the equation (1) Take the derivative e^-8t: dP/dt = -8e^-8t Substitute t = 2 into the derivative dP/dt = -8e^-8(2) dP/dt = -8e^-16 Hence, the rate of change of the population at t = 2 hours is -8e^-16.
We find the rate of change of the population at t = 2 hours by differentiating the population model and substituting t = 2 to get the rate at that specific time.
Derive the second derivative of the function y = e^-8x.
The first step is to find the first derivative, dy/dx = -8e^-8x...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-8e^-8x] Here we apply the chain rule, d²y/dx² = -8(-8)e^-8x = 64e^-8x Therefore, the second derivative of the function y = e^-8x is 64e^-8x.
We use the step-by-step process, starting with the first derivative. Using the chain rule, we differentiate -8e^-8x and simplify the terms to find the second derivative.
Prove: d/dx (e^-16x) = -16e^-16x.
Let’s start using the chain rule: Consider y = e^-16x To differentiate, we apply the chain rule: dy/dx = e^-16x * d/dx(-16x) Since the derivative of -16x is -16, dy/dx = e^-16x * (-16) dy/dx = -16e^-16x Hence proved.
In this step-by-step process, we used the chain rule to differentiate the equation. We replace the inner function with its derivative to derive the equation.
Solve: d/dx (e^-8x/x).
To differentiate the function, we use the quotient rule: d/dx (e^-8x/x) = (d/dx (e^-8x) * x - e^-8x * d/dx(x))/ x² We will substitute d/dx (e^-8x) = -8e^-8x and d/dx(x) = 1 (-8e^-8x * x - e^-8x * 1) / x² = (-8xe^-8x - e^-8x) / x² = e^-8x(-8x - 1) / x² Therefore, d/dx (e^-8x/x) = e^-8x(-8x - 1) / 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. Exponential Function: A mathematical function involving an exponent, typically written as e^x for the natural exponential function. Chain Rule: A rule in calculus for differentiating compositions of functions. Constant Multiplication Rule: A rule stating that the derivative of a constant times a function is the constant times the derivative of the function. Quotient Rule: A method for finding the derivative of a quotient of two functions.

Vikrant is a passionate Mathematics teacher with over 5 years of teaching experience in both conventional and Vedic Maths. His student-focused approach combines clear concepts with motivation and encouragement, helping students build confidence and develo
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