Power Rule for finding Derivatives w/ Step-by-Step Examples

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{d}{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^{'} (x) = \frac{d}{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^{'} (x) = \frac{d}{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:
$f^{'} (x) & = {lim}_{h \to 0} \frac{f (x + h) - f (x)}{h} [2ex] & = {lim}_{h \to 0} \frac{(x + h) - x}{h} [2ex] & = {lim}_{h \to 0} \frac{h}{h} [2ex] & = 1$
This matches the Power Rule formula: $\frac{d}{d x} (x^{1}) = 1 x^{1 - 1} = 1 x^{0} = 1$ .
Inductive step: Assume the Power Rule holds for $n = k$ , i.e., $\frac{d}{d x} (x^{k}) = k x^{k - 1}$ . We need to prove that it also holds for $n = k + 1$ .
Let $f (x) = x^{k + 1}$ . Using the product rule, we get:
$f^{'} (x) = \frac{d}{d x} (x \cdot x^{k}) = x \cdot \frac{d}{d x} (x^{k}) + x^{k} \cdot \frac{d}{d x} (x)$
By the inductive hypothesis and the fact that $\frac{d}{d x} (x) = 1$ , we have:
$f^{'} (x) = x \cdot k x^{k - 1} + x^{k} \cdot 1 = k x^{k} + x^{k} = (k + 1) x^{k}$
This matches the Power Rule formula for $n = k + 1$ : $\frac{d}{d x} (x^{k + 1}) = (k + 1) x^{(k + 1) - 1} = (k + 1) x^{k}$ .

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^{'} (x) = \frac{d}{d x} (\frac{1}{x^{n}}) = \frac{x^{n} \cdot \frac{d}{d x} (1) - 1 \cdot \frac{d}{d x} (x^{n})}{(x^{n})^{2}}

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

f^{'} (x) = \frac{0 - 1 \cdot n x^{n - 1}}{(x^{n})^{2}} = - \frac{n x^{n - 1}}{x^{2 n}} = - n x^{- n - 1}

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

Some Other Power Rules in Calculus

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

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

(x^{2})^{3} = x^{2 \cdot 3} = 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:

\log_{b} (x^{n}) = n \log_{b} (x)

For example:

\log_{2} (x^{3}) = 3 \log_{2} (x)

Power Rule in Differentiation for finding Derivatives