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Power Rule in Derivatives

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  • Difficulty Level : Expert
  • Last Updated : 24 Nov, 2022
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The power rule is a commonly used rule in derivatives. The power rule basically states that the derivative of a variable raised to a power n is n times the variable raised to power n-1. The mathematical formula of power rule can be written as: 

\dfrac{d}{dx}x^n=nx^{n-1} \\\qquad\\

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 a power series with a function’s derivatives.


Find the derivative of 

1. x101

\dfrac{d}{dx}x^n=nx^{n-1}\\\qquad\\ \implies \dfrac{d}{dx}x^{101}=101x^{100}\\\qquad\\

2. 15x6


Power Rule (with rewriting the expression)

From the above equation and example, you now know how to differentiate a variable raised to a power n. 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:

Find the derivative of

1.\ \ x^{\dfrac{-3}{4}}\\\qquad\\ \dfrac{d}{dx}x^{\dfrac{-3}{4}}\ =\ \dfrac{-3}{4}x^{\dfrac{-3}{4}-1}\ =\ \dfrac{-3}{4}x^{\dfrac{-3-4}{4}}\ =\ \dfrac{-3}{4}x^{\dfrac{-7}{4}} \\\qquad\\\qquad\\ 2. \ \ \sqrt{x}\\\qquad\\ \dfrac{d}{dx}\sqrt{x}\ =\ \dfrac{d}{dx}x^{\dfrac{1}{2}}=\dfrac{1}{2}x^{\dfrac{1}{2}-1}\ =\ \dfrac{1}{2}x^{\dfrac{-1}{2}}\ =\ \dfrac{1}{2\sqrt{x}} \\\qquad\\\qquad\\ 3. \ \ \dfrac{1}{\sqrt[3]{x}}\\\qquad\\ \dfrac{d}{dx}\dfrac{1}{\sqrt[3]{x}}\ =\ \dfrac{d}{dx}x^{\dfrac{-1}{3}}\ =\ \dfrac{-1}{3}x^{\dfrac{-1}{3}-1}\ =\ \dfrac{-1}{3}x^{\dfrac{-1-3}{3}}\ =\ \dfrac{-1}{3}x^{\dfrac{-4}{3}}\ =\dfrac{-1}{3\sqrt[3]{x^{4}}} \\\qquad\\\qquad\\ 4. \ \ \sqrt[5]{x^7}\\\qquad\\ \dfrac{d}{dx}\sqrt[5]{x^7}\ =\ \dfrac{d}{dx}x^{\dfrac{7}{5}}\ =\ \dfrac{7}{5}x^{\dfrac{7-5}{5}}\ =\ \dfrac{7}{5}x^{\dfrac{2}{5}} \\\qquad\\\qquad\\

Justifying the Power Rule 


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 binomial theorem we expand (x + △x)n th term

(x+\triangle x)^n\ term\\\qquad\\ \lim\limits_{x\rarr0}\dfrac{(x+\triangle x)^n-x^n}{\triangle x}\\\qquad\\ =\ \lim\limits_{x\rarr0}\dfrac{(\dbinom{n}{0}x^n+\dbinom{n}{1}x^{n-1}\triangle x+\dbinom{n}{2}x^{n-2}\triangle x^2....+\dbinom{n}{n}\triangle x^n)-x^n}{\triangle x}\\\qquad\\ =\ \lim\limits_{x\rarr0}\dfrac{\dbinom{n}{1}x^{n-1}\triangle x+\dbinom{n}{2}x^{n-2}\triangle x^2....+\dbinom{n}{n}\triangle x^n}{\triangle x}\\\qquad\\ = \ \lim\limits_{x\rarr0}\dbinom{n}{0}x^n+\dbinom{n}{1}x^{n-1}+\dbinom{n}{2}x^{n-2}\triangle x....+\dbinom{n}{n}\triangle x^n-1)-x^n\\\qquad\\ = \ \binom{n}{1}x^{n-1}\ =\ nx^{n-1} \\\qquad\\

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

\dfrac{d}{dx}x^n\ =\ nx^{n-1}

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