## What is Power Rule?

The Power Rule is a rule used in calculus for differentiating functions where a variable is **raised to a power,** like ${x}^{5}$. It makes it easier to find the derivative of polynomials and other functions with power terms. The power rule states that to find the derivative of a variable raised to a constant power, you multiply the power by the coefficient and then decrease the power by one.

## Power Rule Formula and Formal Definition

The Power Rule formula is as follows:

$$\frac{\text{d}}{\text{d}x}({x}^{n})=n{x}^{n-1}$$For a function $f(x)={x}^{n}$, where $n$ is a real number, the derivative of $f(x)$ with respect to $x$ is given by:

$${f}^{\prime}(x)=\frac{\text{d}}{\text{d}x}({x}^{n})=n{x}^{n-1}$$## Application of Power Rule

The Power Rule is used when you need to find the derivative of a function that involves a variable raised to a constant power. This rule is particularly useful for **differentiating polynomials,** which are sums of terms with different powers of the variable.

For example, to find the derivative of $f(x)={x}^{3}$, you would apply the Power Rule:

$${f}^{\prime}(x)=\frac{\text{d}}{\text{d}x}({x}^{3})=3{x}^{3-1}=3{x}^{2}$$## Mathematical Proof

There are several ways to prove the Power Rule, including using mathematical induction, the binomial theorem, and the definition of the derivative.

### Power Rule Proof Using Mathematical Induction

We can prove the Power Rule using mathematical induction for positive integer exponents.

Base case: For $n=1$, we have $f(x)={x}^{1}=x$. Using the definition of the derivative, we get:

$${f}^{\prime}(x)\&={lim}_{h\to 0}\frac{f(x+h)-f(x)}{h}\phantom{\rule{0ex}{0ex}}[\mathrm{2ex}]\&={lim}_{h\to 0}\frac{(x+h)-x}{h}\phantom{\rule{0ex}{0ex}}[\mathrm{2ex}]\&={lim}_{h\to 0}\frac{h}{h}\phantom{\rule{0ex}{0ex}}[\mathrm{2ex}]\&=1$$This matches the Power Rule formula: $\frac{\text{d}}{\text{d}x}({x}^{1})=1{x}^{1-1}=1{x}^{0}=1$.

Inductive step: Assume the Power Rule holds for $n=k$, i.e., $\frac{\text{d}}{\text{d}x}({x}^{k})=k{x}^{k-1}$. We need to prove that it also holds for $n=k+1$.

Let $f(x)={x}^{k+1}$. Using the product rule, we get:

$${f}^{\prime}(x)=\frac{\text{d}}{\text{d}x}(x\xb7{x}^{k})=x\xb7\frac{\text{d}}{\text{d}x}({x}^{k})+{x}^{k}\xb7\frac{\text{d}}{\text{d}x}(x)$$By the inductive hypothesis and the fact that $\frac{\text{d}}{\text{d}x}(x)=1$, we have:

$${f}^{\prime}(x)=x\xb7k{x}^{k-1}+{x}^{k}\xb71=k{x}^{k}+{x}^{k}=(k+1){x}^{k}$$This matches the Power Rule formula for $n=k+1$: $\frac{\text{d}}{\text{d}x}({x}^{k+1})=(k+1){x}^{(k+1)-1}=(k+1){x}^{k}$.

Thus, by mathematical induction, the Power Rule holds for all positive integer exponents.

### Power Rule Formula Proof for Negative Integers

To prove the Power Rule for negative integer exponents, we can use the fact that ${x}^{-n}=\frac{1}{{x}^{n}}$ and the Power Rule for positive integer exponents.

Let $f(x)={x}^{-n}$, where $n$ is a positive integer. Using the quotient rule, we get:

$${f}^{\prime}(x)=\frac{\text{d}}{\text{d}x}(\frac{1}{{x}^{n}})=\frac{{x}^{n}\xb7\frac{\text{d}}{\text{d}x}(1)-1\xb7\frac{\text{d}}{\text{d}x}({x}^{n})}{({x}^{n}{)}^{2}}$$Since $\frac{\text{d}}{\text{d}x}(1)=0$ and $\frac{\text{d}}{\text{d}x}({x}^{n})=n{x}^{n-1}$ (by the Power Rule for positive integer exponents), we have:

$${f}^{\prime}(x)=\frac{0-1\xb7n{x}^{n-1}}{({x}^{n}{)}^{2}}=-\frac{n{x}^{n-1}}{{x}^{2n}}=-n{x}^{-n-1}$$This matches the Power Rule formula for negative integer exponents: $\frac{\text{d}}{\text{d}x}({x}^{-n})=-n{x}^{-n-1}$.

## Some Other Power Rules in Calculus

### Power Rule For Exponents: $({x}^{m}{)}^{n}$ = ${x}^{mn}$

This rule states that when raising a power to another power, you can multiply the exponents. For example:

$$({x}^{2}{)}^{3}={x}^{2\xb73}={x}^{6}$$### Power Rule For Logarithms

The Power Rule for Logarithms states that the logarithm of a variable raised to a power is equal to the power multiplied by the logarithm of the variable. In other words:

$${\mathrm{log}}_{b}({x}^{n})=n{\mathrm{log}}_{b}(x)$$For example:

$${\mathrm{log}}_{2}({x}^{3})=3{\mathrm{log}}_{2}(x)$$