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Last updated on September 16, 2025
We use the derivative of 5e^x, which is 5e^x, to understand 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 5e^x in detail.
We now understand the derivative of 5e^x. It is commonly represented as d/dx (5e^x) or (5e^x)', and its value is 5e^x. The function 5e^x has a clearly defined derivative, indicating it is differentiable within its domain.
The key concepts are mentioned below: Exponential Function: (e^x is the base of natural logarithms).
Constant Multiple Rule: A rule for differentiating a constant times a function.
Exponential Rule: The derivative of e^x is e^x.
The derivative of 5e^x can be denoted as d/dx (5e^x) or (5e^x)'.
The formula we use to differentiate 5e^x is: d/dx (5e^x) = 5e^x (or) (5e^x)' = 5e^x
The formula applies to all x in the real number domain.
We can derive the derivative of 5e^x using proofs. To show this, we will use the rules of differentiation. There are several methods we use to prove this, such as:
We will now demonstrate that the differentiation of 5e^x results in 5e^x using the above-mentioned methods:
The derivative of 5e^x can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient.
To find the derivative of 5e^x using the first principle, we will consider f(x) = 5e^x. Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1)
Given that f(x) = 5e^x, we write f(x + h) = 5e^(x + h).
Substituting these into equation (1), f'(x) = limₕ→₀ [5e^(x + h) - 5e^x] / h = 5 limₕ→₀ [e^x (e^h - 1)] / h
We know that e^h - 1 is approximately h for small h, f'(x) = 5e^x limₕ→₀ h / h = 5e^x
Hence, proved.
To prove the differentiation of 5e^x using the constant multiple rule, We use the formula: d/dx (cf(x)) = c d/dx f(x) Let c = 5 and f(x) = e^x. d/dx (5e^x) = 5 d/dx (e^x)
Since the derivative of e^x is e^x, = 5e^x
The exponential rule states that the derivative of e^x is e^x. For the function 5e^x, the derivative is: d/dx (5e^x) = 5 d/dx (e^x) = 5e^x
Thus, the derivative of 5e^x is 5e^x.
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 5e^x.
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 5e^x, it remains the same, as the function is an exponential function with a constant coefficient.
The derivative of 5e^x is always 5e^x, regardless of the value of x, as the exponential function is defined for all real numbers. Since exponential functions do not have asymptotes, the derivative does not have undefined points.
Students frequently make mistakes when differentiating 5e^x. 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 5e^x · e^x
Here, we have f(x) = 5e^x · e^x. Using the product rule, f'(x) = u′v + uv′ In the given equation, u = 5e^x and v = e^x.
Let’s differentiate each term, u′= d/dx (5e^x) = 5e^x v′= d/dx (e^x) = e^x
substituting into the given equation, f'(x) = (5e^x)(e^x) + (5e^x)(e^x)
Let’s simplify terms to get the final answer, f'(x) = 10e^(2x)
Thus, the derivative of the specified function is 10e^(2x).
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 certain population grows exponentially over time and is modeled by the function P(t) = 5e^t, where t is time in years. If t = 3 years, calculate the rate of growth of the population at that time.
We have P(t) = 5e^t (population growth model)...(1)
Now, we will differentiate equation (1) Take the derivative 5e^t: dP/dt = 5e^t
Given t = 3 (substitute this into the derivative) dP/dt = 5e^3
Hence, we get the rate of growth of the population at t = 3 years as 5e^3.
We find the rate of growth of the population at t = 3 years, which shows how rapidly the population increases at that specific time.
Derive the second derivative of the function y = 5e^x.
The first step is to find the first derivative, dy/dx = 5e^x...(1)
Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [5e^x] = 5e^x
Therefore, the second derivative of the function y = 5e^x is 5e^x.
We use the step-by-step process, where we start with the first derivative. Since the derivative of e^x is e^x, and the constant remains, the second derivative is the same as the first.
Prove: d/dx (25e^2x) = 50e^2x.
Let’s start using the chain rule: Consider y = 25e^2x
To differentiate, we use the chain rule: dy/dx = 25 d/dx (e^2x) = 25 * 2e^2x = 50e^2x
Hence proved.
In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replace e^2x with its derivative. As a final step, we simplify to derive the equation.
Solve: d/dx (5e^x/x)
To differentiate the function, we use the quotient rule: d/dx (5e^x/x) = (d/dx (5e^x) · x - 5e^x · d/dx(x))/x²
We will substitute d/dx (5e^x) = 5e^x and d/dx(x) = 1 = (5e^x · x - 5e^x · 1) / x² = (5xe^x - 5e^x) / x² = 5e^x(x - 1) / x²
Therefore, d/dx (5e^x/x) = 5e^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.