Last updated on July 21st, 2025
We use the derivative of x^n, which is n*x^(n-1), as a fundamental tool to understand how polynomial functions change 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 x^n in detail.
We now understand the derivative of x^n. It is commonly represented as d/dx (x^n) or (x^n)', and its value is n*x^(n-1). The function x^n has a clearly defined derivative, indicating it is differentiable within its domain.
The key concepts are mentioned below:
Power Rule: A rule for differentiating functions of the form x^n.
Polynomial Functions: Functions like x^n, where n is a non-negative integer.
The derivative of x^n can be denoted as d/dx (x^n) or (x^n)'. The formula we use to differentiate x^n is: d/dx (x^n) = n*x^(n-1) This formula applies to all x and for any real number n.
We can derive the derivative of x^n using proofs. To show this, we will use the properties of limits 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 x^n results in n*x^(n-1) using the above-mentioned methods:
The derivative of x^n can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient. To find the derivative of x^n
using the first principle, we will consider f(x) = x^n. Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1)
Given that f(x) = x^n, we write f(x + h) = (x + h)^n.
Substituting these into equation (1), f'(x) = limₕ→₀ [(x + h)^n - x^n] / h
Using the Binomial Theorem, expand (x + h)^n: = limₕ→₀ [x^n + n*x^(n-1)*h + ... + h^n - x^n] / h = limₕ→₀ [n*x^(n-1)*h + ... + h^n] / h = limₕ→₀ [n*x^(n-1) + ... + h^(n-1)]
Using limit properties, all terms containing h vanish as h approaches 0, leaving: f'(x) = n*x^(n-1) Hence, proved.
To prove the differentiation of x^n using the binomial theorem, We use the expansion: (x + h)^n = x^n + n*x^(n-1)*h + ... + h^n
By focusing on the terms that involve h, when we divide by h and take the limit as h approaches 0, we find that only the first-order term, n*x^(n-1), remains, yielding the derivative.
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 x^n.
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 x^n, 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 (continuing for higher-order derivatives).
When n = 0, the derivative is undefined because x^0 = 1, which is a constant with a derivative of 0. When n = 1, the derivative of x^1 = 1*x^(1-1), which is 1.
Students frequently make mistakes when differentiating x^n. 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 x^3 * e^x
Here, we have f(x) = x^3 * e^x. Using the product rule, f'(x) = u′v + uv′ In the given equation, u = x^3 and v = e^x. Let’s differentiate each term, u′ = d/dx (x^3) = 3x^2 v′ = d/dx (e^x) = e^x
Substituting into the given equation, f'(x) = (3x^2) * (e^x) + (x^3) * (e^x)
Let’s simplify terms to get the final answer, f'(x) = 3x^2 * e^x + x^3 * e^x f'(x) = e^x (3x^2 + x^3)
Thus, the derivative of the specified function is e^x (3x^2 + x^3).
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 rocket is being launched, and its height is given by h(x) = x^3 + 5x^2 meters, where x is time in seconds. Find the rate of change of height when x = 2 seconds.
We have h(x) = x^3 + 5x^2 (height of the rocket)...(1)
Now, we will differentiate the equation (1)
Take the derivative: h'(x) = 3x^2 + 10x
Given x = 2 (substitute this into the derivative)
h'(2) = 3*(2)^2 + 10*(2) = 3*4 + 20 = 12 + 20 = 32
Hence, the rate of change of height when x = 2 seconds is 32 meters/second.
We find the rate of change of height at x = 2 seconds as 32 meters/second, meaning the rocket's height is increasing at this rate at that moment in time.
Derive the second derivative of the function y = x^4.
The first step is to find the first derivative, dy/dx = 4x^3...(1)
Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [4x^3] = 4 * 3x^2 = 12x^2
Therefore, the second derivative of the function y = x^4 is 12x^2.
We use the step-by-step process, starting with the first derivative. We apply the power rule again to find the second derivative, resulting in 12x^2.
Prove: d/dx (x^2 + x^3) = 2x + 3x^2.
Let’s apply the power rule: Consider y = x^2 + x^3.
To differentiate, we take each term separately: dy/dx = d/dx (x^2) + d/dx (x^3) = 2x + 3x^2
Hence proved.
In this step-by-step process, we differentiate each term separately using the power rule. We then add the derivatives to derive the equation.
Solve: d/dx (x^2/x)
To differentiate the function, we simplify first: x^2/x = x
Now take the derivative: d/dx (x) = 1 Therefore, d/dx (x^2/x) = 1.
In this process, we simplify the given function first and then differentiate the simplified form, resulting in the derivative of 1.
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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