Last updated on July 15th, 2025
We use the derivative of e^-4x, which is -4e^-4x, as a tool for understanding how the exponential function changes in response to a slight change in x. Derivatives help us calculate profit or loss in real-life situations. We will now discuss the derivative of e^-4x in detail.
We now understand the derivative of e^-4x. It is commonly represented as d/dx (e^-4x) or (e^-4x)', and its value is -4e^-4x. The function e^-4x has a clearly defined derivative, indicating it is differentiable for all real numbers. The key concepts are mentioned below: Exponential Function: (e^-4x is an exponential function). Chain Rule: Rule for differentiating composite functions (since e^-4x is a composite function of e^x).
The derivative of e^-4x can be denoted as d/dx (e^-4x) or (e^-4x)'. The formula we use to differentiate e^-4x is: d/dx (e^-4x) = -4e^-4x The formula applies to all real numbers x.
We can derive the derivative of e^-4x using proofs. To show this, we will use the chain rule along with the rules of differentiation. There are several methods we use to prove this, such as: Using Chain Rule Using Chain Rule To prove the differentiation of e^-4x using the chain rule, We use the formula: Let u = -4x, then y = e^u. By chain rule: dy/dx = dy/du * du/dx dy/du = d/du (e^u) = e^u du/dx = d/dx (-4x) = -4 Substituting these into the chain rule, dy/dx = e^u * (-4) Since u = -4x, dy/dx = e^-4x * (-4) dy/dx = -4e^-4x Hence, proved.
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 help us understand functions like e^-4x better. 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^-4x, 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).
The function e^-4x is always differentiable for all real numbers x. When x = 0, the derivative of e^-4x is -4e^0, which equals -4.
Students frequently make mistakes when differentiating e^-4x. 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^-4x * e^3x
Here, we have f(x) = e^-4x * e^3x. Using the product rule, f'(x) = u′v + uv′ In the given equation, u = e^-4x and v = e^3x. Let’s differentiate each term, u′ = d/dx (e^-4x) = -4e^-4x v′ = d/dx (e^3x) = 3e^3x Substituting into the given equation, f'(x) = (-4e^-4x) * (e^3x) + (e^-4x) * (3e^3x) Let’s simplify terms to get the final answer, f'(x) = -4e^-x + 3e^-x f'(x) = (-4 + 3)e^-x f'(x) = -e^-x Thus, the derivative of the specified function is -e^-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 radioactive substance decays over time, and its quantity is represented by the function Q = e^-4x, where Q is the quantity remaining and x is the time in hours. Calculate the rate of decay when x = 1 hour.
We have Q = e^-4x (rate of decay)...(1) Now, we will differentiate the equation (1) Take the derivative of e^-4x: dQ/dx = -4e^-4x Given x = 1 (substitute this into the derivative) dQ/dx = -4e^-4(1) dQ/dx = -4e^-4 dQ/dx ≈ -0.0733 (since e^-4 ≈ 0.0183) Hence, we get the rate of decay at x = 1 hour as approximately -0.0733.
We find the rate of decay at x = 1 hour to be approximately -0.0733, which means that at this point, the quantity is decreasing at this rate.
Derive the second derivative of the function Q = e^-4x.
The first step is to find the first derivative, dQ/dx = -4e^-4x...(1) Now we will differentiate equation (1) to get the second derivative: d²Q/dx² = d/dx [-4e^-4x] d²Q/dx² = -4 * d/dx [e^-4x] d²Q/dx² = -4 * (-4e^-4x) d²Q/dx² = 16e^-4x Therefore, the second derivative of the function Q = e^-4x is 16e^-4x.
We use the step-by-step process, where we start with the first derivative. Using the chain rule, we differentiate again. We then simplify the terms to find the final answer.
Prove: d/dx (e^-8x) = -8e^-8x.
Let’s start using the chain rule: Consider y = e^-8x To differentiate, we use the chain rule: dy/dx = e^-8x * d/dx (-8x) Since the derivative of -8x is -8, dy/dx = e^-8x * (-8) Substituting y = e^-8x, d/dx (e^-8x) = -8e^-8x Hence proved.
In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replace -8x with its derivative. As a final step, we substitute y = e^-8x to derive the equation.
Solve: d/dx (e^-4x/x)
To differentiate the function, we use the quotient rule: d/dx (e^-4x/x) = (d/dx (e^-4x) * x - e^-4x * d/dx(x))/ x² We will substitute d/dx (e^-4x) = -4e^-4x and d/dx (x) = 1 = (-4e^-4x * x - e^-4x * 1) / x² = (-4xe^-4x - e^-4x) / x² = -e^-4x (4x + 1) / x² Therefore, d/dx (e^-4x/x) = -e^-4x (4x + 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 function of the form e^kx, where k is a constant, such as e^-4x. Chain Rule: A rule for differentiating composite functions by differentiating the outer function and multiplying by the derivative of the inner function. Quotient Rule: A method for differentiating a function that is the quotient of two other functions. Exponential Decay: A decrease in quantity that follows an exponential function, such as e^-4x.
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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