Last updated on August 5th, 2025
We use the derivative of e^(x+3), which is e^(x+3), as a measuring tool for 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 talk about the derivative of e^(x+3) in detail.
We now understand the derivative of e^(x+3). It is commonly represented as d/dx (e^(x+3)) or (e^(x+3))', and its value is e^(x+3). The function e^(x+3) has a clearly defined derivative, indicating it is differentiable for all real numbers.
The key concepts are mentioned below:
Exponential Function: e^(x+3) is an exponential function where the base is Euler's number.
Chain Rule: Rule used for differentiating composite functions like e^(x+3).
Derivative of e^x: The derivative of the natural exponential function e^x is itself, e^x.
The derivative of e^(x+3) can be denoted as d/dx (e^(x+3)) or (e^(x+3))'. The formula we use to differentiate e^(x+3) is: d/dx (e^(x+3)) = e^(x+3)
The formula applies to all x.
We can derive the derivative of e^(x+3) using proofs. To show this, we will use the rules of differentiation. There are several methods we use to prove this, such as: U
To prove the differentiation of e^(x+3) using the chain rule, Let u = x + 3, which makes e^(x+3) = e^u. Then, d/dx (e^(x+3)) = d/du (e^u) * du/dx
The derivative of e^u with respect to u is e^u and the derivative of u with respect to x is 1.
So, d/dx (e^(x+3)) = e^u * 1 = e^(x+3).
The derivative of e^(x+3) can also be proved using the First Principle, which expresses the derivative as the limit of the difference quotient.
Consider f(x) = e^(x+3). Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h = limₕ→₀ [e^(x+h+3) - e^(x+3)] / h = limₕ→₀ e^(x+3) [e^h - 1] / h
Using the limit formula limₕ→₀ (e^h - 1)/h = 1, f'(x) = e^(x+3) * 1 = e^(x+3).
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 e^(x+3).
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^(x+3), 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.
The derivative of e^(x+3) is always e^(x+3) regardless of x, as the exponential function is defined for all real numbers. However, at x = -3, e^(x+3) simplifies to e^0, which is 1.
Similarly, at any point, the derivative is simply the function value itself.
Students frequently make mistakes when differentiating e^(x+3). 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^(x+3) * ln(x))
Here, we have f(x) = e^(x+3) * ln(x).
Using the product rule, f'(x) = u′v + uv′ In the given equation, u = e^(x+3) and v = ln(x).
Let’s differentiate each term, u′ = d/dx (e^(x+3)) = e^(x+3) v′ = d/dx (ln(x)) = 1/x
Substituting into the given equation, f'(x) = (e^(x+3))(1/x) + (e^(x+3))(ln(x))
Let’s simplify terms to get the final answer, f'(x) = e^(x+3)/x + e^(x+3)ln(x)
Thus, the derivative of the specified function is e^(x+3)/x + e^(x+3)ln(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 models its revenue growth with the function R(x) = e^(x+3) where x is time in years. Find the rate of revenue growth at x = 2 years.
We have R(x) = e^(x+3) (revenue growth function)...(1)
Now, we will differentiate the equation (1)
Take the derivative e^(x+3): dR/dx = e^(x+3) Given x = 2, substitute this into the derivative, dR/dx = e^(2+3) = e^5
Hence, we get the rate of revenue growth at x = 2 years as e^5.
We find the rate of revenue growth at x = 2 years by substituting x = 2 into the derivative of the given function. This gives us the rate of change of revenue at that specific time.
Derive the second derivative of the function y = e^(x+3).
The first step is to find the first derivative, dy/dx = e^(x+3)...(1)
Now, we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [e^(x+3)]
Since the derivative of e^(x+3) is e^(x+3), d²y/dx² = e^(x+3)
Therefore, the second derivative of the function y = e^(x+3) is e^(x+3).
Using the step-by-step process, we start with the first derivative. Since the derivative of an exponential function remains the same, the second derivative is also e^(x+3).
Prove: d/dx (e^(2x+6)) = 2e^(2x+6).
Let’s start using the chain rule: Consider y = e^(2x+6) We use the chain rule: dy/dx = d/du (e^u) * du/dx where u = 2x+6
The derivative of e^u with respect to u is e^u and the derivative of u with respect to x is 2. dy/dx = e^(2x+6) * 2
Thus, d/dx (e^(2x+6)) = 2e^(2x+6).
Hence proved.
In this step-by-step process, we use the chain rule to differentiate the equation. We differentiate the inner function and multiply it with the derivative of the outer function to derive the equation.
Solve: d/dx (e^(x+3)/x)
To differentiate the function, we use the quotient rule: d/dx (e^(x+3)/x) = (d/dx (e^(x+3)).x - e^(x+3).d/dx(x))/x²
We will substitute d/dx (e^(x+3)) = e^(x+3) and d/dx (x) = 1 = (e^(x+3).x - e^(x+3))/x² = e^(x+3)(x - 1)/x²
Therefore, d/dx (e^(x+3)/x) = e^(x+3)(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.
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