Question

If
$y = x^{n}$
Find
$\frac{d ^{n}y }{dx ^{n} }$
​

Answer 1

Given,

$\longrightarrow y=x^n$

The first derivative of $y$ is,

$\longrightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}(x^n)$

$\longrightarrow \dfrac{dy}{dx}=nx^{n-1}$

$\longrightarrow \dfrac{dy}{dx}=\dfrac{n(n-1)!}{(n-1)!}\cdot x^{n-1}$

$\longrightarrow \dfrac{dy}{dx}=\dfrac{n!}{(n-1)!}\cdot x^{n-1}$

$\longrightarrow \dfrac{dy}{dx}=\ ^n\!P_1\ x^{n-1}$

The second derivative of $y$ is,

$\longrightarrow \dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(\dfrac{dy}{dx}\right)$

$\longrightarrow \dfrac{d^2y}{dx^2}=\dfrac{d}{dx}(nx^{n-1})$

$\longrightarrow \dfrac{d^2y}{dx^2}=n\cdot\dfrac{d}{dx}(x^{n-1})$

$\longrightarrow \dfrac{d^2y}{dx^2}=n(n-1)x^{n-2}$

$\longrightarrow \dfrac{d^2y}{dx^2}=\dfrac{n(n-1)(n-2)!}{(n-2)!}\cdot x^{n-2}$

$\longrightarrow \dfrac{d^2y}{dx^2}=\dfrac{n!}{(n-2)!}\cdot x^{n-2}$

$\longrightarrow \dfrac{d^2y}{dx^2}=\ ^n\!P_2\ x^{n-2}$

Assume that, for $n=k\in\mathbb{N},$ $k^{th}$ derivative of $y$ is,

$\longrightarrow \dfrac{d^ky}{dx^k}=\ ^n\!P_k\ x^{n-k}\quad\quad\dots(1)$

Let $n=k+1$ such that, $(k+1)^{th}$ derivative of $y$ is,

$\longrightarrow \dfrac{d^{k+1}y}{dx^{k+1}}=\ ^n\!P_{k+1}\ x^{n-(k+1)}\quad\quad\dots(2)$

Let's check whether it's true.

$\longrightarrow \dfrac{d^{k+1}y}{dx^{k+1}}=\dfrac{d}{dx}\left(\dfrac{d^ky}{dx^k}\right)$

From (1),

$\longrightarrow \dfrac{d^{k+1}y}{dx^{k+1}}=\dfrac{d}{dx}\left(^n\!P_k\ x^{n-k}\right)$

$\longrightarrow \dfrac{d^{k+1}y}{dx^{k+1}}=\,^n\!P_k\cdot\dfrac{d}{dx}\left(x^{n-k}\right)$

$\longrightarrow \dfrac{d^{k+1}y}{dx^{k+1}}=\dfrac{n!}{(n-k)!}\cdot(n-k)x^{n-k-1}$

$\longrightarrow \dfrac{d^{k+1}y}{dx^{k+1}}=\dfrac{n!}{(n-k-1)!}\cdot x^{n-k-1}$

$\longrightarrow \dfrac{d^{k+1}y}{dx^{k+1}}=\dfrac{n!}{(n-(k+1))!}\cdot x^{n-(k+1)}$

$\longrightarrow \dfrac{d^{k+1}y}{dx^{k+1}}=\ ^n\!P_{k+1}\ x^{n-(k+1)}$

Hence (2) is true whenever (1) is true.

Therefore, $n^{th}$ derivative of $y$ is,

$\longrightarrow \dfrac{d^ny}{dx^n}=\ ^n\!P_n\ x^{n-n}$

$\longrightarrow\underline{\underline{\dfrac{d^ny}{dx^n}=n!}}$