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Last updated on October 11, 2025
We use the derivative of 8e^x, which is 8e^x, as a tool to measure how the 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 8e^x in detail.
We now understand the derivative of 8ex. It is commonly represented as d/dx (8ex) or (8ex)', and its value is 8ex. The function 8e^x has a clearly defined derivative, indicating it is differentiable within its domain.
The key concepts are mentioned below:
Exponential Function: (ex) is a fundamental mathematical constant.
Constant Multiplication Rule: The derivative of a constant multiplied by a function.
Derivative of ex: The derivative of ex is ex itself.
The derivative of 8ex can be denoted as d/dx (8ex) or (8ex)'.
The formula we use to differentiate 8ex is: d/dx (8ex) = 8ex (or) (8ex)' = 8ex
The formula applies to all x.
We can derive the derivative of 8ex using proofs. To show this, we will use the properties of exponents along with the rules of differentiation.
There are several methods we use to prove this, such as:
We will now demonstrate that the differentiation of 8ex results in 8ex using the above-mentioned methods:
The derivative of 8ex can be found by applying the constant multiplication rule, which states that the derivative of a constant times a function is the constant times the derivative of the function. If f(x) = 8ex, then f'(x) = 8 * d/dx (ex) We know the derivative of ex is ex, f'(x) = 8 * ex Thus, f'(x) = 8ex.
To prove the differentiation of 8ex using the chain rule, We consider the function as a composition of two functions: 8 times ex. Let f(x) = 8 and g(x) = ex. The derivative of the product of these functions is given by: f'(x) = f'(x)g(x) + f(x)g'(x) Since f'(x) = 0 (as 8 is a constant), f'(x) = 0 * ex + 8 * ex f'(x) = 8ex. 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 make it easier to understand functions like 8ex.
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 8ex, 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 derivative of 8ex is defined for all x since the exponential function is continuous everywhere.
For any constant value of x, the derivative remains 8ex.
Students frequently make mistakes when differentiating 8ex.
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 (8e^x * ln(x))
Here, we have f(x) = 8e^x * ln(x). Using the product rule, f'(x) = u′v + uv′ In the given equation, u = 8e^x and v = ln(x). Let’s differentiate each term, u′ = d/dx (8e^x) = 8e^x v′ = d/dx (ln(x)) = 1/x Substituting into the given equation, f'(x) = (8e^x)(ln(x)) + (8e^x)(1/x) Let’s simplify terms to get the final answer, f'(x) = 8e^x ln(x) + 8e^x/x. Thus, the derivative of the specified function is 8e^x ln(x) + 8e^x/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 new factory produces goods at a rate represented by the function y = 8e^x, where y represents the production rate and x represents time in hours. If x = 1 hour, calculate the rate of change of production.
We have y = 8e^x (production rate)...(1) Now, we will differentiate the equation (1). Take the derivative 8e^x: dy/dx = 8e^x Given x = 1 (substitute this into the derivative), dy/dx = 8e^1 = 8e. Hence, the rate of change of production at x = 1 hour is 8e.
We find the rate of change of production at x = 1 hour by substituting the value into the derivative.
This provides the production rate change at that specific time.
Derive the second derivative of the function y = 8e^x.
The first step is to find the first derivative, dy/dx = 8e^x...(1) Now, we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [8e^x] As the derivative of 8e^x is 8e^x, d²y/dx² = 8e^x. Therefore, the second derivative of the function y = 8e^x is 8e^x.
We use the step-by-step process, where we start with the first derivative.
Since the derivative of 8ex remains 8ex, the second derivative is the same.
Prove: d/dx (4e^x) = 4e^x.
Consider y = 4e^x. To differentiate, we apply the constant multiplication rule: dy/dx = 4 * d/dx (e^x) Since the derivative of e^x is e^x, dy/dx = 4 * e^x Thus, d/dx (4e^x) = 4e^x. Hence, proved.
In this step-by-step process, we used the constant multiplication rule to differentiate the equation.
Then, we replace ex with its derivative to derive the equation.
Solve: d/dx (8e^x/x)
To differentiate the function, we use the quotient rule: d/dx (8e^x/x) = (d/dx (8e^x) * x - 8e^x * d/dx(x)) / x² We will substitute d/dx (8e^x) = 8e^x and d/dx (x) = 1, = (8e^x * x - 8e^x * 1) / x² = (8e^x x - 8e^x) / x² = 8e^x (x - 1) / x² Therefore, d/dx (8e^x/x) = 8e^x (x - 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.
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.
: He loves to play the quiz with kids through algebra to make kids love it.