Last updated on July 21st, 2025
We use the derivative of u^x, where u is a constant, to understand how the function u^x changes with respect to x. Derivatives are crucial in calculating changes in various scenarios, including growth and decay in real-life situations. We will now explore the derivative of u^x in detail.
We now understand the derivative of u^x. It is commonly represented as d/dx (u^x) or (u^x)', and its value is u^x ln(u). The function u^x has a well-defined derivative, indicating it is differentiable.
The key concepts are mentioned below:
Exponential Function: u^x represents an exponential function.
Logarithmic Function: ln(u) represents the natural logarithm of u.
Chain Rule: Used for differentiating functions involving compositions, like u^x.
The derivative of u^x can be denoted as d/dx (u^x) or (u^x)'. The formula we use to differentiate u^x is: d/dx (u^x) = u^x ln(u)
The formula applies to all x, where u is a positive constant.
We can derive the derivative of u^x using proofs. To show this, we will use the properties of logarithms and the rules of differentiation. There are several methods we use to prove this, such as:
We will now demonstrate that the differentiation of u^x results in u^x ln(u) using the above-mentioned methods:
To find the derivative of u^x using logarithmic differentiation, we take the natural logarithm of both sides. Consider y = u^x. Taking natural log on both sides gives us ln(y) = x ln(u).
Differentiating both sides with respect to x gives us 1/y dy/dx = ln(u). Thus, dy/dx = y ln(u) = u^x ln(u).
Hence, proved.
To prove the differentiation of u^x using the chain rule, We use the fact that u^x can be expressed as e^(x ln(u)).
Let y = e^(x ln(u)). Then, dy/dx = e^(x ln(u)) (d/dx (x ln(u))). Since d/dx (x ln(u)) = ln(u), we have: dy/dx = e^(x ln(u)) ln(u) = u^x ln(u).
Using Exponential Properties
We can also use the exponential properties to derive the derivative of u^x. Consider f(x) = u^x.
Using the property of exponential functions, we express u^x as e^(x ln(u)).
Differentiating with respect to x, we have f'(x) = e^(x ln(u)) ln(u). Thus, f'(x) = u^x ln(u).
Hence, the derivative of u^x is u^x ln(u).
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 u^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 u^x, 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 the base u is 1, the derivative is zero because 1^x is a constant function. When the base u is e, the derivative of e^x is e^x ln(e), which simplifies to e^x.
Students frequently make mistakes when differentiating u^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 (3^x · 5^x)
Here, we have f(x) = 3^x · 5^x. Using the product rule, f'(x) = u′v + uv′ In the given equation, u = 3^x and v = 5^x.
Let’s differentiate each term, u′= d/dx (3^x) = 3^x ln(3) v′= d/dx (5^x) = 5^x ln(5)
Substituting into the given equation, f'(x) = (3^x ln(3))(5^x) + (3^x)(5^x ln(5))
Let’s simplify terms to get the final answer, f'(x) = 3^x 5^x ln(3) + 3^x 5^x ln(5)
Thus, the derivative of the specified function is 3^x 5^x (ln(3) + ln(5)).
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’s revenue is modeled by the function R(x) = 2^x dollars, where x is the number of years since the company’s inception. Calculate the rate of change of revenue when x = 3.
We have R(x) = 2^x (revenue function)...(1)
Now, we will differentiate the equation (1)
Take the derivative of 2^x: dR/dx = 2^x ln(2)
Substituting x = 3 into the derivative: dR/dx = 2^3 ln(2) dR/dx = 8 ln(2)
Hence, the rate of change of revenue when x = 3 is 8 ln(2) dollars per year.
We find the rate of change of revenue at x = 3 as 8 ln(2) dollars per year, indicating how revenue increases at that specific year.
Derive the second derivative of the function y = 4^x.
The first step is to find the first derivative, dy/dx = 4^x ln(4)...(1)
Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [4^x ln(4)] d²y/dx² = 4^x (ln(4))^2
Therefore, the second derivative of the function y = 4^x is 4^x (ln(4))^2.
We use the step-by-step process, where we start with the first derivative. Then, we differentiate it again to find the second derivative using the exponential function properties.
Prove: d/dx ((7^x)^2) = 2 · 7^x · 7^x ln(7).
Let’s start using the chain rule: Consider y = (7^x)^2
To differentiate, we use the chain rule: dy/dx = 2 · 7^x · d/dx (7^x)
Since the derivative of 7^x is 7^x ln(7), dy/dx = 2 · 7^x · 7^x ln(7)
Substituting y = (7^x)^2, d/dx ((7^x)^2) = 2 · 7^x · 7^x ln(7)
Hence proved.
In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replace 7^x with its derivative. As a final step, we substitute y = (7^x)^2 to derive the equation.
Solve: d/dx ((x^2) · (9^x))
To differentiate the function, we use the product rule: d/dx ((x^2) · (9^x)) = (d/dx (x^2) · 9^x + x^2 · d/dx (9^x))
We will substitute d/dx (x^2) = 2x and d/dx (9^x) = 9^x ln(9) = (2x · 9^x + x^2 · 9^x ln(9)) = 2x 9^x + x^2 9^x ln(9)
Therefore, d/dx ((x^2) · (9^x)) = 9^x (2x + x^2 ln(9))
In this process, we differentiate the given function using the product 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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