Power Rule

The power rule is a basic rule in calculus used to find the derivative of expressions like xⁿ.

Rule:

Since differentiation is a linear operation on the space of differentiable functions, polynomials can also be differentiated using this rule. The power rule underlies the Taylor series as it relates to a power series with a function's derivatives.

Example 1: Find the derivative of x¹⁰¹.

Answer:

As \dfrac{d}{dx}x^n=nx^{n-1}\\

\implies \dfrac{d}{dx}x^{101}=101x^{100}\\\qquad\\

Example 2: Find the derivative of 15x⁶.

Answer:

As \dfrac{d}{dx}x^n=nx^{n-1}

\dfrac{d(15x^6)}{dx}=15(6x^{6-1})=90x^5

Power Rule for Non-Integers

The point to be noted is that n can also be fractional, and so the variable could have exponents, and these exponents are real numbers. For better understanding, check the following examples:

Example: Find the derivative of x^{\frac{-3}{4}}

Answer:

\text{Let } f(x) = x^{\frac{-3}{4}}\\ \Rightarrow f'(x) = \frac{d}{dx}x^{\frac{-3}{4}}\\ \Rightarrow f'(x) =\frac{-3}{4}x^{\frac{-3}{4}-1}\\ \Rightarrow f'(x) =\frac{-3}{4}x^{\frac{-3-4}{4}}\\ \Rightarrow f'(x) =\frac{-3}{4}x^{\frac{-7}{4}} 

Example: Find the derivative of √x.

Answer:

\text{Let } f(x) = \sqrt{x}\\ \Rightarrow f'(x) = \frac{d}{dx}\sqrt{x}\ = \frac{d}{dx}x^{\frac{1}{2}}\\ \Rightarrow f'(x) =\frac{1}{2}x^{\frac{1}{2}-1} = \frac{1}{2}x^{\frac{-1}{2}}\\ \Rightarrow f'(x) = \frac{1}{2\sqrt{x}}

Derivation of Power Rule

We can derive the formula for the power rule using two methods, which are as follows:

Using the Principle of Mathematical Induction

The Power Rule states that if f(x) = xⁿ, where n is a positive integer, then f'(x) = nxⁿ⁻¹..

Base Case

Let n=1. Then f(x) = x.

and f'(x) = 1, which is equal to the derivative of x.

Thus, the base case is true.

Inductive Hypothesis

Let us assume that the Power Rule holds true for n=k, where k is an arbitrary positive integer. 

Therefore, if f(x) = x^k, then f'(x) = kx^k-1.

Inductive Step

We need to show that the Power Rule holds for n=k+1. 

Let f(x) = x^k+1 = x × x^k.

Differentiate using the Product Rule, we get:

f'(x) = x^k + x × kx^k-1 

⇒ f'(x) = x^k + kx^k 

⇒ f'(x) = (k+1)x^k

Thus by induction, the power rule holds true for all natural numbers.

Using Binomial Theorems

Using the definition of derivative we can write

\dfrac{d}{dx}x^n\ as\ \lim\limits_{x\rarr0}\dfrac{(x+\triangle x)^n-x^n}{\triangle x}\\\qquad\\  

By using the binomial theorem we expand (x + △x) ^nth term

(x+\triangle x)^n\ term\\\qquad\\\lim\limits_{\triangle x \to 0}\dfrac{(x+\triangle x)^n-x^n}{\triangle x}\\\qquad\\=\ \lim\limits_{\triangle x \to 0}\dfrac{(\dbinom{n}{0}x^n+\dbinom{n}{1}x^{n-1}\triangle x+\dbinom{n}{2}x^{n-2}\triangle x^2+\cdots+\dbinom{n}{n}\triangle x^n)-x^n}{\triangle x}\\\qquad\\=\ \lim\limits_{\triangle x \to 0}\dfrac{\dbinom{n}{1}x^{n-1}\triangle x+\dbinom{n}{2}x^{n-2}\triangle x^2+\cdots+\dbinom{n}{n}\triangle x^n}{\triangle x}\\\qquad\\=\ \lim\limits_{\triangle x \to 0}\left(\dbinom{n}{1}x^{n-1}+\dbinom{n}{2}x^{n-2}\triangle x+\cdots+\dbinom{n}{n}\triangle x^{n-1}\right)\\\qquad\\=\ \dbinom{n}{1}x^{n-1}\ =\ nx^{n-1}\\\qquad\\

Only the first term remained, as it does not contain an △ x term; hence,

\dfrac{d}{dx}x^n\ =\ nx^{n-1} 

Applications of Power Rule

If we have a polynomial function f(x) = a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0,      we can differentiate it by taking the derivative of each term using the power rule and adding the results.

Example: Find the derivative of {3x^4- 2x^3+5x^2 -7x+1}.

Answer:

f'(x) = (d/dx)(3x^4) - (d/dx)(2x^3) + (d/dx)(5x^2) - (d/dx)(7x) + (d/dx)(1) 

⇒f′(x) = 12x³−6x²+10x−7+0

So the derivative of f(x) is f′(x) = 12x³ − 6x² + 10x − 7.

Other Power Rules in Calculus

There are various other power rules used in calculus that are used to solve various problems. Some of the various power rules in calculus are,

Power Rule Integration

Power rule in integration is helpful for finding the integral of expressions that are given as xⁿ, where n is a real number and n ≠ -1. The formula for the integration power rule is,

∫xⁿ dx = xⁿ⁺¹/(n + 1) + C

where n ≠ -1. We can understand this rule using the example discussed below.

(i) ∫10x⁹ dx 

= 10(x⁹⁺¹)/(9+1) + C 

= 10x¹⁰/10 + C 

= x¹⁰ + C

(ii) ∫x^-3 dx

= x^-3+1/(-3+1) + C 

= -x - 2/2 + C 

= -1/2x² + C

Power Rule Exponents

Power rule in exponents is used when we have to find the power of the exponents that are given as,

(x^m)ⁿ = x^mn

We can understand this rule using the example discussed below.

• (x²)⁴ = x^2×4 = x⁸
• (2^-3)^-3 = 2^-3×-3 = 2⁹

Power Rule Logarithms

The power rule in logarithms is used to solve the power of any logarithmic function such as,

log_n(a)^b = b.log_n(a)

We can understand this rule using the example discussed below.

• log₈x³ = 3log₈x

Related Articles

• Applications of Power Rule
• Laws of Exponents

Solved Problems

Problem 1: Find the derivative of f(x) = x⁵.

Solution:

Using the power rule, we have:

f'(x) = 5x^(5-1) = 5x⁴

Problem 2: Find the derivative of {\frac{1}{\sqrt[3]{x}}}.

Solution:

\text{Let } f(x) = \frac{1}{\sqrt[3]{x}}\\ \Rightarrow f'(x) = \frac{d}{dx}\frac{1}{\sqrt[3]{x}}\ =\ \frac{d}{dx}x^{\frac{-1}{3}}\\ \Rightarrow f'(x) = \frac{-1}{3}x^{\frac{-1}{3}-1}\\ \Rightarrow f'(x) =\ \frac{-1}{3}x^{\frac{-1-3}{3}}\\ \Rightarrow f'(x) =\ \frac{-1}{3}x^{\frac{-4}{3}}\\ \Rightarrow f'(x) =\frac{-1}{3\sqrt[3]{x^{4}}}

Problem 3: Find the derivative of {{\sqrt[5]{x^7}}}   .

Solution:

\text{Let } f(x) = \sqrt[5]{x^7}\\ \Rightarrow f'(x) = \frac{d}{dx}\sqrt[5]{x^7}\ \\ \Rightarrow f'(x) = \frac{d}{dx}x^{\frac{7}{5}}\\ \Rightarrow f'(x) = \frac{7}{5}x^{\frac{7-5}{5}}\\ \Rightarrow f'(x) = \frac{7}{5}x^{\frac{2}{5}}

Problem 4: Find the derivative of h(x) = x^-2/3.

Solution: 

Thus, h'(x) = (1/3)(-2)x^-2-1 = (-2/3)x^-3 

Therefore, the derivative of h(x) = x^-2/3  is -2x^-3/3.

Problem 5: Find the derivative of k(x) = (5x² + 3x)⁴.

Solution: 

Using the chain rule and power rule together we get,

 k'(x) = 4(5x^2 + 3x)^{4-1}\times  \frac{d}{dx}(5x^2 + 3x)        

\Rightarrow k'(x) =  4(5x^2 + 3x)^{4-1}\times (10x + 3)

Therefore, the derivative of k(x) = (5x² + 3x)⁴ is  4(5x² + 3x)³(10x + 3).