In calculus, the power rule is used to differentiate functions of the form f ( x ) = x r {\displaystyle f(x)=x^{r}} , whenever r {\displaystyle r} is a real number. 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.
Statement Let f {\displaystyle f} be a function satisfying f ( x ) = x r {\displaystyle f(x)=x^{r}} for all x {\displaystyle x} , where r ∈ R {\displaystyle r\in \mathbb {R} } . Then,
f ′ ( x ) = r x r − 1 . {\displaystyle f'(x)=rx^{r-1}\,.}
The power rule for integration states that
∫ x r d x = x r + 1 r + 1 + C {\displaystyle \int \!x^{r}\,dx={\frac {x^{r+1}}{r+1}}+C}
for any real number r ≠ − 1 {\displaystyle r\neq -1} . It can be derived by inverting the power rule for differentiation. In this equation C is any constant.
Proofs
Proof for real exponents Let f ( x ) = x r {\displaystyle f(x)=x^{r}} , where r {\displaystyle r} is any real number. If f ( x ) = e x {\displaystyle f(x)=e^{x}} , then ln ( f ( x ) ) = x {\displaystyle \ln(f(x))=x} , where ln {\displaystyle \ln } is the natural logarithm function, or f ′ ( x ) = f ( x ) = e x {\displaystyle f'(x)=f(x)=e^{x}} , as was required. Therefore, applying the chain rule to f ( x ) = e r ln x {\displaystyle f(x)=e^{r\ln x}} , we see that
f ′ ( x ) = r x e r ln x = r x x r {\displaystyle f'(x)={\frac {r}{x}}e^{r\ln x}={\frac {r}{x}}x^{r}}
which simplifies to r x r − 1 {\displaystyle rx^{r-1}} . When x < 0 {\displaystyle x<0} , we may use the same definition with x r = ( ( − 1 ) ( − x ) ) r = ( − 1 ) r ( − x ) r {\displaystyle x^{r}=((-1)(-x))^{r}=(-1)^{r}(-x)^{r}} , where we now have − x > 0 {\displaystyle -x>0} . This necessarily leads to the same result. Note that because ( − 1 ) r {\displaystyle (-1)^{r}} does not have a conventional definition when r {\displaystyle r} is not a rational number, irrational power functions are not well defined for negative bases. In addition, as rational powers of −1 with even denominators (in lowest terms) are not real numbers, these expressions are only real valued for rational powers with odd denominators (in lowest terms). Finally, whenever the function is differentiable at x = 0 {\displaystyle x=0} , the defining limit for the derivative is:
lim h → 0 h r − 0 r h {\displaystyle \lim _{h\to 0}{\frac {h^{r}-0^{r}}{h}}}
which yields 0 only when r {\displaystyle r} is a rational number with odd denominator (in lowest terms) and r > 1 {\displaystyle r>1} , and 1 when r = 1 {\displaystyle r=1} . For all other values of r {\displaystyle r} , the expression h r {\displaystyle h^{r}} is not well-defined for h < 0 {\displaystyle h<0} , as was covered above, or is not a real number, so the limit does not exist as a real-valued derivative. For the two cases that do exist, the values agree with the value of the existing power rule at 0, so no exception need be made. The exclusion of the expression 0 0 {\displaystyle 0^{0}} (the case x = 0 {\displaystyle x=0} ) from our scheme of exponentiation is due to the fact that the function f ( x , y ) = x y {\displaystyle f(x,y)=x^{y}} has no limit at (0,0), since x 0 {\displaystyle x^{0}} approaches 1 as x approaches 0, while 0 y {\displaystyle 0^{y}} approaches 0 as y approaches 0. Thus, it would be problematic to ascribe any particular value to it, as the value would contradict one of the two cases, dependent on the application. It is traditionally left undefined.
Proofs for integer exponents
Proof by induction (natural numbers) Let n ∈ N {\displaystyle n\in \mathbb {N} } . It is required to prove that d d x x n = n x n − 1 . {\displaystyle {\frac {d}{dx}}x^{n}=nx^{n-1}.} The base case may be when n = 0 {\displaystyle n=0} or 1 {\displaystyle 1} , depending on how the set of natural numbers is defined. When n = 0 {\displaystyle n=0} , d d x x 0 = d d x ( 1 ) = lim h → 0 1 − 1 h = lim h → 0 0 h = 0 = 0 x 0 − 1 . {\displaystyle {\frac {d}{dx}}x^{0}={\frac {d}{dx}}(1)=\lim _{h\to 0}{\frac {1-1}{h}}=\lim _{h\to 0}{\frac {0}{h}}=0=0x^{0-1}.}
When n = 1 {\displaystyle n=1} , d d x x 1 = lim h → 0 ( x + h ) − x h = lim h → 0 h h = 1 = 1 x 1 − 1 . {\displaystyle {\frac {d}{dx}}x^{1}=\lim _{h\to 0}{\frac {(x+h)-x}{h}}=\lim _{h\to 0}{\frac {h}{h}}=1=1x^{1-1}.} Therefore, the base case holds either way. Suppose the statement holds for some natural number k, i.e. d d x x k = k x k − 1 . {\displaystyle {\frac {d}{dx}}x^{k}=kx^{k-1}.}
When n = k + 1 {\displaystyle n=k+1} , d d x x k + 1 = d d x ( x k ⋅ x ) = x k ⋅ d d x x + x ⋅ d d x x k = x k + x ⋅ k x k − 1 = x k + k x k = ( k + 1 ) x k = ( k + 1 ) x ( k + 1 ) − 1 {\displaystyle {\frac {d}{dx}}x^{k+1}={\frac {d}{dx}}(x^{k}\cdot x)=x^{k}\cdot {\frac {d}{dx}}x+x\cdot {\frac {d}{dx}}x^{k}=x^{k}+x\cdot kx^{k-1}=x^{k}+kx^{k}=(k+1)x^{k}=(k+1)x^{(k+1)-1}} By the principle of mathematical induction, the statement is true for all natural numbers n.
Proof by binomial theorem (natural number) Let y = x n {\displaystyle y=x^{n}} , where n ∈ N {\displaystyle n\in \mathbb {N} } . Then, d y d x = lim h → 0 ( x + h ) n − x n h = lim h → 0 1 h [ x n + ( n 1 ) x n − 1 h + ( n 2 ) x n − 2 h 2 + ⋯ + ( n n ) h n − x n ] = lim h → 0 [ ( n 1 ) x n − 1 + ( n 2 ) x n − 2 h + ⋯ + ( n n ) h n − 1 ] = n x n − 1 {\displaystyle {\begin{aligned}{\frac {dy}{dx}}&=\lim _{h\to 0}{\frac {(x+h)^{n}-x^{n}}{h}}\\[4pt]&=\lim _{h\to 0}{\frac {1}{h}}\left[x^{n}+{\binom {n}{1}}x^{n-1}h+{\binom {n}{2}}x^{n-2}h^{2}+\dots +{\binom {n}{n}}h^{n}-x^{n}\right]\\[4pt]&=\lim _{h\to 0}\left[{\binom {n}{1}}x^{n-1}+{\binom {n}{2}}x^{n-2}h+\dots +{\binom {n}{n}}h^{n-1}\right]\\[4pt]&=nx^{n-1}\end{aligned}}}
Since n choose 1 is equal to n, and the rest of the terms all contain h, which is 0, the rest of the terms cancel. This proof only works for natural numbers as the binomial theorem only works for natural numbers.
Proof by Geometric Series (natural number) Let y = x n {\displaystyle y=x^{n}} , where n ∈ N {\displaystyle n\in \mathbb {N} }
and a , b ∈ R {\displaystyle a,b\in \mathbb {R} } . Then, d y d x | x = a = lim b → a b n − a n b − a = lim b → a [ a n − 1 + a n − 2 b + a n − 3 b 2 + ⋯ + b n − 1 ] = n a n − 1 {\displaystyle {\begin{aligned}\left.{\frac {dy}{dx}}\right|_{x=a}&=\lim _{b\to a}{\frac {b^{n}-a^{n}}{b-a}}\\[4pt]&=\lim _{b\to a}\left[a^{n-1}+a^{n-2}b+a^{n-3}b^{2}+\dots +b^{n-1}\right]\\[4pt]&=na^{n-1}\end{aligned}}}
Generalization to negative integer exponents For a negative integer n, let n = − m {\displaystyle n=-m} so that m is a positive integer. Using the reciprocal rule, d d x x n = d d x ( 1 x m ) = − d d x x m ( x m ) 2 = − m x m − 1 x 2 m = − m x − m − 1 = n x n − 1 . {\displaystyle {\frac {d}{dx}}x^{n}={\frac {d}{dx}}\left({\frac {1}{x^{m}}}\right)={\frac {-{\frac {d}{dx}}x^{m}}{(x^{m})^{2}}}=-{\frac {mx^{m-1}}{x^{2m}}}=-mx^{-m-1}=nx^{n-1}.} In conclusion, for any integer n {\displaystyle n} , d d x x n = n x n − 1 . {\displaystyle {\frac {d}{dx}}x^{n}=nx^{n-1}.}
Generalization to rational exponents Upon proving that the power rule holds for integer exponents, the rule can be extended to rational exponents.
Proof by chain rule This proof is composed of two steps that involve the use of the chain rule for differentiation.
Let y = x r = x 1 n {\displaystyle y=x^{r}=x^{\frac {1}{n}}} , where n ∈ N + {\displaystyle n\in \mathbb {N} ^{+}} . Then y n = x {\displaystyle y^{n}=x} . By the chain rule, n y n − 1 ⋅ d y d x = 1 {\displaystyle ny^{n-1}\cdot {\frac {dy}{dx}}=1} . Solving for d y d x {\displaystyle {\frac {dy}{dx}}} , d y d x = 1 n y n − 1 = 1 n ( x 1 n ) n − 1 = 1 n x 1 − 1 n = 1 n x 1 n − 1 = r x r − 1 {\displaystyle {\frac {dy}{dx}}={\frac {1}{ny^{n-1}}}={\frac {1}{n\left(x^{\frac {1}{n}}\right)^{n-1}}}={\frac {1}{nx^{1-{\frac {1}{n}}}}}={\frac {1}{n}}x^{{\frac {1}{n}}-1}=rx^{r-1}} Thus, the power rule applies for rational exponents of the form 1 / n {\displaystyle 1/n} , where n {\displaystyle n} is a nonzero natural number. This can be generalized to rational exponents of the form p / q {\displaystyle p/q} by applying the power rule for integer exponents using the chain rule, as shown in the next step. Let y = x r = x p / q {\displaystyle y=x^{r}=x^{p/q}} , where p ∈ Z , q ∈ N + , {\displaystyle p\in \mathbb {Z} ,q\in \mathbb {N} ^{+},} so that r ∈ Q {\displaystyle r\in \mathbb {Q} } . By the chain rule, d y d x = d d x ( x 1 q ) p = p ( x 1 q ) p − 1 ⋅ 1 q x 1 q − 1 = p q x p / q − 1 = r x r − 1 {\displaystyle {\frac {dy}{dx}}={\frac {d}{dx}}\left(x^{\frac {1}{q}}\right)^{p}=p\left(x^{\frac {1}{q}}\right)^{p-1}\cdot {\frac {1}{q}}x^{{\frac {1}{q}}-1}={\frac {p}{q}}x^{p/q-1}=rx^{r-1}}
From the above results, we can conclude that when r {\displaystyle r} is a rational number, d d x x r = r x r − 1 . {\displaystyle {\frac {d}{dx}}x^{r}=rx^{r-1}.}
Proof by implicit differentiation A more straightforward generalization of the power rule to rational exponents makes use of implicit differentiation. Let y = x r = x p / q {\displaystyle y=x^{r}=x^{p/q}} , where p , q ∈ Z {\displaystyle p,q\in \mathbb {Z} } so that r ∈ Q {\displaystyle r\in \mathbb {Q} } . Then, y q = x p {\displaystyle y^{q}=x^{p}} Differentiating both sides of the equation with respect to x {\displaystyle x} , q y q − 1 ⋅ d y d x = p x p − 1 {\displaystyle qy^{q-1}\cdot {\frac {dy}{dx}}=px^{p-1}} Solving for d y d x {\displaystyle {\frac {dy}{dx}}} , d y d x = p x p − 1 q y q − 1 . {\displaystyle {\frac {dy}{dx}}={\frac {px^{p-1}}{qy^{q-1}}}.} Since y = x p / q {\displaystyle y=x^{p/q}} , d d x x p / q = p x p − 1 q x p − p / q . {\displaystyle {\frac {d}{dx}}x^{p/q}={\frac {px^{p-1}}{qx^{p-p/q}}}.} Applying laws of exponents, d d x x p / q = p q x p − 1 x − p + p / q = p q x p / q − 1 . {\displaystyle {\frac {d}{dx}}x^{p/q}={\frac {p}{q}}x^{p-1}x^{-p+p/q}={\frac {p}{q}}x^{p/q-1}.} Thus, letting r = p q {\displaystyle r={\frac {p}{q}}} , we can conclude that d d x x r = r x r − 1 {\displaystyle {\frac {d}{dx}}x^{r}=rx^{r-1}} when r {\displaystyle r} is a rational number.
Proof by Geometric Series (Positive Rationals) This proof uses specific algebraic formulas to evaluate the definitional form of the derivative exactly for positive rational powers. If y = x p q {\displaystyle y=x^{\frac {p}{q}}} , where p , q ∈ N {\displaystyle p,q\in \mathbb {N} }
and a , b ∈ R {\displaystyle a,b\in \mathbb {R} } . Then,