Proof $\dfrac{d}{dx}{x^n}$ formula

Formula

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

$x$ is a variable and it is raised to power of $n$. The varying quantity in a function form is written as $x^{\displaystyle n}$. The derivative of $x^{\displaystyle n}$ with respect to $x$ is used as a formula in differential calculus to differentiate the functions which are in power form.

The differentiation of $x^{\displaystyle n}$ law can be derived in differential calculus by using fundamental method of finding derivative of function with respect to $x$.

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

$f{(x)} \,=\, x^{\displaystyle n}$, then $f{(x+h)} \,=\, {(x+h)}^{\displaystyle n}$

Substitute the functions in the formula

Replace the differentiation of a function formula by replacing the values of the functions $f{(x)}$ and $f{(x+h)}$.

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

Simplify the function by taking out common factors

$x^{\displaystyle n}$ is a term in the numerator and the same term can be taken out from the term ${(x+h)}^{\displaystyle n}$.

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

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

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

Apply expansion of Binomial Theorem

The term ${\Big(1+\dfrac{h}{x}\Big)}^{\displaystyle n}$ is in the form binomial theorem. So, it can be expanded by applying Binomial theorem.

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

In this case, $\dfrac{h}{x}$ is there instead of $x$. So, replace $x$ by $\dfrac{h}{x}$ in the expansion of the binomial theorem.

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

Simplify the function to obtain derivative of the function

There is a lot of scope for simplification of this expression and it helps us to get the differentiation of $x^{\displaystyle n}$ easily in the next few steps.

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

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

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

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

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

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

Get quotient of $x^{\displaystyle n}$ by $x$ by using quotient rule of exponents.

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

Get derivative of the function

Substitute $h$ is equal to zero and simplify the expression to get the derivative of $x$ is raised to the power of $n$ with respect to $x$ in differential calculus.

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

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

$\implies \dfrac{d}{dx}{x^{\displaystyle n}}$ $\,=\,$ $x^{{\displaystyle n}-1}$ $\times$ $[n]$

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

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Apr 17, 2019
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