Question

y=x³.Find out double differenciation​

Answer 1

$\large\underline{\sf{Given- }}$

$\rm \: y = {x}^{3}$

$\large\underline{\sf{To\:Find - }}$

$\rm \: \dfrac{ {d}^{2} y}{d {x}^{2} }$

$\large\underline{\sf{Solution-}}$

Given function is

$\rm \: y = {x}^{3}$

On differentiating both sides w. r. t. x, we get

$\rm \: \dfrac{d}{dx}y = \dfrac{d}{dx}{x}^{3}$

We know,

$\boxed{\rm{  \:\dfrac{d}{dx} {x}^{n} = {nx}^{n - 1} \: }}$

So, using this result, we get

$\rm \: \dfrac{dy}{dx} = {3x}^{2}$

On differentiating both sides w. r. t. x, we get

$\rm \: \dfrac{d}{dx}\dfrac{dy}{dx} = \dfrac{d}{dx} {3x}^{2}$

$\rm \: \dfrac{ {d}^{2} y}{d {x}^{2} } = 3\dfrac{d}{dx} {x}^{2}$

$\rm \: \dfrac{ {d}^{2} y}{d {x}^{2} } = 3 \times 2x$

$\rm\implies \:\rm \: \dfrac{ {d}^{2} y}{d {x}^{2} } = 6x$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$

Answer 2

Solution:

Given That:

$\rm \longrightarrow y = {x}^{3}$

Differentiating wrt x, we get:

$\rm \longrightarrow \dfrac{dy}{dx} = \dfrac{d}{dx}( {x}^{3} )$

We know that:

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

Therefore:

$\rm \longrightarrow \dfrac{dy}{dx} = 3 {x}^{3 - 1}$

$\rm \longrightarrow \dfrac{dy}{dx} = 3 {x}^{2}$

Differentiating again wrt x, we get:

$\rm \longrightarrow \dfrac{ {d}^{2} y}{d {x}^{2} } = \dfrac{d}{dx} (3 {x}^{2})$

$\rm \longrightarrow \dfrac{ {d}^{2} y}{d {x}^{2} } = 3\dfrac{d}{dx} ( {x}^{2})$

$\rm \longrightarrow \dfrac{ {d}^{2} y}{d {x}^{2} } = 3 \times 2x$

$\rm \longrightarrow \dfrac{ {d}^{2} y}{d {x}^{2} } = 6x$

★ Which is our required answer.

Learn More:

$\begin{gathered}\boxed{\begin{array}{c|c}\bf f(x)&\bf\dfrac{d}{dx}f(x)\\ \\ \frac{\qquad\qquad}{}&\frac{\qquad\qquad}{}\\ \sf k&\sf0\\ \\ \sf sin(x)&\sf cos(x)\\ \\ \sf cos(x)&\sf-sin(x)\\ \\ \sf tan(x)&\sf{sec}^{2}(x)\\ \\ \sf cot(x)&\sf-{cosec}^{2}(x)\\ \\ \sf sec(x)&\sf sec(x)tan(x)\\ \\ \sf cosec(x)&\sf-cosec(x)cot(x)\\ \\ \sf\sqrt{x}&\sf\dfrac{1}{2\sqrt{x}}\\ \\ \sf log(x)&\sf\dfrac{1}{x}\\ \\ \sf{e}^{x}&\sf{e}^{x}\end{array}}\\ \end{gathered}$