Proof of Power Rule of Derivatives

$x$ is a variable and $n$ is a constant. The power rule of differentiation can be derived from first principle in differential calculus to find the derivative of exponential function $x^{\displaystyle n}$ with respect to $x$.

Write Derivative of function in Limit form

Write the derivative of a function in limits form by the definition of the derivative.

$\dfrac{d}{dx}{\, f(x)}$ $\,=\,$ $\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{f(x+\Delta x)-f(x)}{\Delta x}}$

It can be written in terms of $h$ by taking $\Delta x$ equals to $h$.

$\implies$ $\dfrac{d}{dx}{\, f(x)}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{f(x+h)-f(x)}{h}}$

Take $f{(x)} \,=\, x^{\displaystyle n}$, then $f{(x+h)} \,=\, {(x+h)}^{\displaystyle n}$. Now, substitute them in the above formula to get started deriving this formula from first principle.

$\implies \dfrac{d}{dx}{\, (x^{\displaystyle n})}$ $\,=\,$ $\large \displaystyle \lim_{h \,\to 0} \normalsize \dfrac{{(x+h)}^{\displaystyle n}-x^{\displaystyle n}}{h}$

Simplify the Algebraic function

$=\,\,\,$ $\large \displaystyle \lim_{h \,\to 0} \normalsize \dfrac{{(x+h)}^{\displaystyle n}-x^{\displaystyle n}}{h}$

$=\,\,\,$ $\large \displaystyle \lim_{h \,\to 0} \normalsize \dfrac{x^{\displaystyle n}{\Big(1+\dfrac{h}{x}\Big)}^{\displaystyle n}-x^{\displaystyle n}}{h}$

Look at the algebraic expression in the numerator. An exponential function $x^{\displaystyle n}$ can be taken common from both terms.

$=\,\,\,$ $\large \displaystyle \lim_{h \,\to 0} \normalsize \dfrac{x^{\displaystyle n}\Bigg[{\Big(1+\dfrac{h}{x}\Big)}^{\displaystyle n}-1\Bigg]}{h}$

Now, factorize this algebraic function as two functions.

$=\,\,\,$ $\large \displaystyle \lim_{h \,\to 0} \normalsize \Bigg(\dfrac{x^{\displaystyle n}}{h} \times {\Bigg[{\Big(1+\dfrac{h}{x}\Big)}^{\displaystyle n}-1\Bigg]\Bigg)}$

Evaluate Binomial function by Binomial Theorem

The function ${\Big(1+\dfrac{h}{x}\Big)}^{\displaystyle n}$ can be expanded by the binomial theorem.

According to Binomial Theorem, we know that

${(1+x)}^{\displaystyle n}$ $\,=\,$ $1$ $+$ $\dfrac{nx}{1!}$ $+$ $\dfrac{n(n-1)}{2!}x^2$ $+$ $\dfrac{n(n-1)(n-2)}{3!}x^3$ $+$ $\cdots$

Now, replace $x$ by the fraction $\dfrac{h}{x}$ for expanding the function ${\Big(1+\dfrac{h}{x}\Big)}^{\displaystyle n}$.

$=\,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0} \normalsize \Bigg(\dfrac{x^{\displaystyle n}}{h}$ $\times$ $\Bigg[1$ $+$ $\dfrac{n}{1!}{\Big(\dfrac{h}{x}\Big)}$ $+$ $\dfrac{n(n-1)}{2!}{\Big(\dfrac{h}{x}\Big)}^2$ $+$ $\dfrac{n(n-1)(n-2)}{3!}{\Big(\dfrac{h}{x}\Big)}^3$ $+$ $\cdots -1\Bigg]\Bigg)$

Now, simplify the algebraic expression.

$=\,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0} \normalsize \Bigg(\dfrac{x^{\displaystyle n}}{h}$ $\times$ $\Bigg[\require{cancel} \cancel{1}$ $+$ $\dfrac{n}{1!}{\Big(\dfrac{h}{x}\Big)}$ $+$ $\dfrac{n(n-1)}{2!}{\Big(\dfrac{h}{x}\Big)}^2$ $+$ $\dfrac{n(n-1)(n-2)}{3!}{\Big(\dfrac{h}{x}\Big)}^3$ $+$ $\cdots -\require{cancel} \cancel{1}\Bigg]\Bigg)$

$=\,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0} \normalsize \Bigg(\dfrac{x^{\displaystyle n}}{h}$ $\times$ $\Bigg[\dfrac{n}{1!}{\Big(\dfrac{h}{x}\Big)}$ $+$ $\dfrac{n(n-1)}{2!}{\Big(\dfrac{h}{x}\Big)}^2$ $+$ $\dfrac{n(n-1)(n-2)}{3!}{\Big(\dfrac{h}{x}\Big)}^3$ $+$ $\cdots \Bigg]\Bigg)$

The fraction $\dfrac{h}{x}$ is a common factor in all the terms of the infinite series. So, take it out common from all the terms.

$=\,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0} \normalsize \Bigg(\dfrac{x^{\displaystyle n}}{h} $ $\times$ ${\Big(\dfrac{h}{x}\Big)}\Bigg[\dfrac{n}{1!}$ $+$ $\dfrac{n(n-1)}{2!}{\Big(\dfrac{h}{x}\Big)}$ $+$ $\dfrac{n(n-1)(n-2)}{3!}{\Big(\dfrac{h}{x}\Big)}^2$ $+$ $\cdots \Bigg]\Bigg)$

$=\,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0} \normalsize \Bigg(\dfrac{x^{\displaystyle n} \times h}{h \times x} $ $\times$ $\Bigg[\dfrac{n}{1!}$ $+$ $\dfrac{n(n-1)}{2!}{\Big(\dfrac{h}{x}\Big)}$ $+$ $\dfrac{n(n-1)(n-2)}{3!}{\Big(\dfrac{h}{x}\Big)}^2$ $+$ $\cdots \Bigg]\Bigg)$

$=\,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0} \normalsize \require{cancel} \Bigg(\dfrac{x^{\displaystyle n} \times \cancel{h}}{\cancel{h} \times x} $ $\times$ $\Bigg[\dfrac{n}{1!}$ $+$ $\dfrac{n(n-1)}{2!}{\Big(\dfrac{h}{x}\Big)}$ $+$ $\dfrac{n(n-1)(n-2)}{3!}{\Big(\dfrac{h}{x}\Big)}^2$ $+$ $\cdots \Bigg]\Bigg)$

$=\,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0} \normalsize \Bigg(\dfrac{x^{\displaystyle n}}{x} $ $\times$ $\Bigg[\dfrac{n}{1!}$ $+$ $\dfrac{n(n-1)}{2!}{\Big(\dfrac{h}{x}\Big)}$ $+$ $\dfrac{n(n-1)(n-2)}{3!}{\Big(\dfrac{h}{x}\Big)}^2$ $+$ $\cdots \Bigg]\Bigg)$

According to quotient rule of exponents with same base rule, find the quotient by diving $x^{\displaystyle n}$ with $x$.

$=\,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0} \normalsize \Bigg(x^{{\displaystyle n}-1}$ $\times$ $\Bigg[\dfrac{n}{1!}$ $+$ $\dfrac{n(n-1)}{2!}{\Big(\dfrac{h}{x}\Big)}$ $+$ $\dfrac{n(n-1)(n-2)}{3!}{\Big(\dfrac{h}{x}\Big)}^2$ $+$ $\cdots \Bigg]\Bigg)$

Evaluate the Limit of the infinite series

Now, find the limit of the infinite series as h approaches zero by the direct substitution method.

$=\,\,\,$ $x^{{\displaystyle n}-1}$ $\times$ $\Bigg[\dfrac{n}{1!}$ $+$ $\dfrac{n(n-1)}{2!}{\Big(\dfrac{0}{x}\Big)}$ $+$ $\dfrac{n(n-1)(n-2)}{3!}{\Big(\dfrac{0}{x}\Big)}^2$ $+$ $\cdots \Bigg]$

$=\,\,\,$ $x^{{\displaystyle n}-1}$ $\times$ $\Big[\dfrac{n}{1}$ $+$ $0$ $+$ $0$ $+$ $\cdots \Big]$

$=\,\,\,$ $x^{{\displaystyle n}-1}$ $\times$ $[n]$

$\,\,\, \therefore \,\,\,\,\,\, \dfrac{d}{dx}{\, (x^{\displaystyle n})}$ $\,=\,$ $nx^{{\displaystyle n}-1}$

Thus, the power rule of derivatives is derived in differential calculus by first principle.