In calculus , the power rule is used for differential functions of the form whenever a is a real number. Since differentiation is a linear operation in place of discriminant functions, polynomials can also be differentiated using this rule. The power rule outlines the Taylor series as it relates to a power series with derivatives of a function.

f(x) = x^rR

**Statement of power law**

Let the work be satisfactory for all , together . Then,

{\displaystyle f:\mathbb {R} \mapsto \mathbb {R} }{\displaystyle f(x)=x^{r}}xr\in {\mathbb {R}}

{\displaystyle f'(x)=rx^{r-1}\,.}

The power law for integration states that

\int\! x^r \, dx=\frac{x^{r+1}}{r+1}+C

For any real number . This can be obtained by reversing the power rule for differentiation.

r \neq -1

**Evidence**

**Proof for real exponentiation**

To begin, we must choose a working definition of the value of k , where is any real number. Although it is possible to define value as the limit of a sequence of rational powers that reach an irrational power whenever we encounter such a power, or as at least the upper bound of the set of rational powers that are less than a given power. In the U.S., such a definition is not liable to discrimination. It is therefore preferable to use a functional definition, which is usually taken to have all values of , where is the natural exponential function and is is Euler’s number . [1] [2] First, we can show that the derivative of .

f(x) = x^rr

x^r = \exp(r\ln x) = e^{r\ln x}x>0\exp ef(x)=e^{x}f'(x) = e^x

If , then then , where is the natural logarithmic function, the inverse function of the exponential function, as demonstrated by Euler. [3] Since the latter two functions . are equal for all values of k , their derivatives are also equal, whenever a derivative exists, we have, by the chain rule ,

f(x)=e^{x}\ln (f(x)) = x\ln x>0

\frac{1}{f(x)}\cdot f'(x) = 1

Or , as required. So, applying the chain rule , we see that

f'(x)=f(x)=e^xf(x) = e^{r\ln x}

f'(x)=\frac{r}{x}e^{r\ln x}=\frac{r}{x}x^r

that simplifies . *rx ^{r-1}*

When , we can use the same definition as we have now . This essentially leads to the same result. Note that because there is no conventional definition when a rational number is not, irrational power functions are not well defined for negative bases. Furthermore, as rational powers of -1 are not real numbers with even denominators (in lowest terms), these expressions are only of real value for rational powers with odd denominators (in lowest terms). *x < 0*

x^r = ((-1)(-x))^r = (-1)^r(-x)^r-x > 0(-1)^rr

Finally, whenever the function is differentiable , the limit defined for the derivative is: *x=0*

\lim_{h\to 0} \frac{h^r - 0^r}{h}

which yields 0 only if the odd denominator is a rational number (in lowest terms) and , and 1 when r = 1. For all other values of r, the expression is not well defined , as above was told, or is not a real number, so the limit does not exist as a real-valued derivative. For the two cases that exist, the values agree to the value of the existing power rule at 0, so no exceptions need to be made.

r > 1h^rh < 0

The exclusion of the expression (case x = 0) from our scheme of exponentiation is due to the fact that the function (0,0) has no limit, because as x approaches 0, approaches 1, while as -As y approaches 0, it approaches 0. Thus, assigning it to a particular value would be problematic, as the value would be the opposite of one of two cases, depending on the application. It has traditionally been left undefined.

0^{0}f(x, y) = x^yx^{0}0^y

**Proof for non-zero integer exponentiation**

**Proof by induction (positive integer)**

Let *n* be a positive integer. it is necessary to prove that

{\displaystyle {\frac {d}{dx}}x^{n}=nx^{n-1}.}

When , therefore, the base case stands.

n=1{\displaystyle {\frac {d}{dx}}x^{1}=\lim _{h\to 0}{\frac {(x+h)-x}{h}}=1=1x^{1-1}.}

Let the statement hold *for some positive integer k* , i.e.

{\displaystyle {\frac {d}{dx}}x^{k}=kx^{k-1}.}

when ,

n=k+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}}

According to the principle of mathematical induction, this statement is true for all positive integers *n .*

**Proof by Binomial Theorem (Positive Integer)**

Lashkar , where is

y=x^n{\displaystyle n\in \mathbb {N} }

Then

{\displaystyle {\frac {dy}{dx}}=\lim _{h\to 0}{\frac {(x+h)^{n}-x^{n}}{h}}} {\displaystyle =\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}+...+{\binom {n}{n}}h^{n}-x^{n}\right]}

{\displaystyle =\lim _{h\to 0}\left[{\binom {n}{1}}x^{n-1}+{\binom {n}{2}}x^{n-2}h+...+{\binom {n}{n}}h^{n-1}\right]}

{\displaystyle =nx^{n-1}}

**Normalization of Negative Integer Exponents**

For a negative integer n , let so that m is a positive integer. Using the reciprocal rule , *n = – m*

{\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}.}

Finally, for any non-zero integer ,

n{\displaystyle {\frac {d}{dx}}x^{n}=nx^{n-1}.}

**Generalization to Rational Exponents**

On proving that the power law holds for integer exponents, the rule can be extended to rational exponents.

**case-by-case generalization**

1. Come on , where is it

{\displaystyle y=x^{\frac {1}{n}}=x^{m}}n\in \mathbb{N}

Then

{\displaystyle y^{n}=x}

By the chain rule we get

{\displaystyle ny^{n-1}\cdot {\frac {dy}{dx}}=1}

thus,

{\displaystyle {\frac {dy}{dx}}={\frac {1}{ny^{n-1}}}={\frac {1}{n\left(x^{\frac {1}{n}}\right)^{n-1}}}={\frac {1}{n}}x^{{\frac {1}{n}}-1}=mx^{m-1}}

2. Let , where is , so that

{\displaystyle y=x^{\frac {n}{m}}=x^{p}}{\displaystyle m,n\in \mathbb {N} }{\displaystyle p\in \mathbb {Q} ^{+}}

by chain rule ,

{\displaystyle {\frac {dy}{dx}}={\frac {d}{dx}}\left(x^{\frac {1}{m}}\right)^{n}=n\left(x^{\frac {1}{m}}\right)^{n-1}\cdot {\frac {1}{m}}x^{{\frac {1}{m}}-1}={\frac {n}{m}}x^{{\frac {n}{m}}-1}=px^{p-1}}

3. Let , where is and

{\displaystyle y=x^{q}}{\displaystyle q=-p}{\displaystyle p\in \mathbb {Q} ^{+}}

Using the chain rule and the reciprocal rule , we have

{\displaystyle {\frac {dy}{dx}}={\frac {d}{dx}}\left({\frac {1}{x}}\right)^{p}=p\left({\frac {1}{x}}\right)^{p-1}\cdot \left(-{\frac {1}{x^{2}}}\right)=-px^{-p-1}=qx^{q-1}}

From the above results, we can conclude that when r is a rational number ,

{\displaystyle {\frac {d}{dx}}x^{r}=rx^{r-1}.}

proof by implicit differentiation

A more direct generalization of the power law to rational exponents uses implicit differentiation.

Lashkar , where is so that .

{\displaystyle y=x^{r}=x^{p/q}}{\displaystyle p,q\in \mathbb {Z} }{\displaystyle r\in \mathbb {Q} }

Then

{\displaystyle y^{q}=x^{p}}\\

{\displaystyle qy^{q-1}{\frac {dy}{dx}}=px^{p-1}}

to solve for , *dy/dx*

{\displaystyle {\frac {dy}{dx}}={\frac {px^{p-1}}{qy^{q-1}}}.}

Since , *y = x ^{p/q}*

{\displaystyle {\frac {d}{dx}}x^{p/q}={\frac {px^{p-1}}{qx^{p-p/q}}}.}

Applying the laws of exponentiation,

{\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, giving , we can conclude that when a is a rational number.

{\displaystyle r=p/q}{\displaystyle {\frac {d}{dx}}x^{r}=rx^{r-1}}r

**History**

The power rule for integrals was first demonstrated in a geometric form for all positive integer values in the early 17th century by the Italian mathematician Bonaventura Cavalieri and in the mid-17th century by mathematicians Pierre de Fermat , Evangelista Torricelli , Gilles de Roberval , For All Rational Powers by John Wallis and Blaise Pascal , each acting independently. At that time, they were the treatise on determining the area between the graph and the horizontal axis of a rational power function. However, in retrospect, it is considered the first general theorem of calculus to be discovered. [4] Isaac Newton ‘s power law for resolution{\displaystyle {\displaystyle n}}and was obtained, each independently, for rational power functions in the mid-17th century by Gottfried Wilhelm Leibniz , who then used it as an inverse operation to derive power rules for integrals. It introduces related theorems in the traditional way in modern basic calculus textbooks, where differentiation rules usually precede integration rules. [5]

Although both men stated that their rules, demonstrated only for rational quantities, work for all real powers, neither asked for such proof, as the application of the theory at the time did not pertain to such foreign power functions. , and the question of convergence of infinite series was still unclear.

The unique case was solved in the mid-17th century by the Flemish Jesuit and mathematician Gregoire de Saint-Vincent and his student Alphonse Antonio de Sarsa , who showed that the related definite integral, *r = -1*

\int_1^x \frac{1}{t}\, dt

Representing the area between the rectangular hyperbola and the x-axis, was a logarithmic function whose basis was eventually discovered as the transcendental number e . The modern notation for the value of this definite integral is the natural logarithm.

xy=1\ln(x)

**Generalization**

**Complex power work**

If we consider functions of the form where is any complex number and a slit is a complex number in the complex plane that intersects a branch point of 0 and any branch connected to it, and we use the traditional multiplicative definition , So it is straightforward to show that, on each branch of the complex logarithm, the same logic used above gives a similar result:

f(z) = z^cczz^c := \exp(c\ln z)f'(z) = \frac{c}{z}\exp(c\ln z)

Furthermore, if is a positive integer, there is no need for branch deduction: one can define , or define positive integral complex powers via complex multiplication, and show that for all complex , derivative and binomial theorems . from the definition of.

cf(0)=0f'(z) = cz^{c-1}z

However, due to the multi-valued nature of complex power functions for non-integer exponents, one must be careful in specifying the branch of the complex logarithm being used. Furthermore, no matter which branch is used , the function is not differentiable at 0 if it is not a positive integer. *c*