Question

second order derivative of x^x​

Answer 1

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

Given function is

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

On taking log on both sides, we get

$\rm \: logy = log({x}^{x})$

can be rewritten as

$\rm \: logy = x \: logx$

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

$\rm \: \dfrac{d}{dx}logy = \dfrac{d}{dx}( x \: logx)$

$\rm \: \dfrac{1}{y}\dfrac{dy}{dx} = x\dfrac{d}{dx}logx + logx\dfrac{d}{dx}x$

$\rm \: \dfrac{1}{y}\dfrac{dy}{dx} = x \times \frac{1}{x} + logx \times 1$

$\rm \: \dfrac{1}{y}\dfrac{dy}{dx} = 1 + logx$

$\rm\implies \:\dfrac{dy}{dx} = y(1 + logx)$

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

$\rm \: \dfrac{d}{dx}\bigg(\dfrac{dy}{dx}\bigg) = y\dfrac{d}{dx}(1 + logx) + (1 + logx)\dfrac{d}{dx}y$

$\rm \: \dfrac{ {d}^{2} y}{d {x}^{2} } = y \times \dfrac{1}{x} + (1 + logx)\dfrac{dy}{dx}$

$\rm \: \dfrac{ {d}^{2} y}{d {x}^{2} } = \dfrac{ {x}^{x} }{x} + (1 + logx) \times y(1 + logx)$

$\rm \: \dfrac{ {d}^{2} y}{d {x}^{2} } = \dfrac{ {x}^{x} }{x} + {x}^{x} {(1 + logx)}^{2}$

$\rm \: \dfrac{ {d}^{2} y}{d {x}^{2} } = {x}^{x - 1} + {x}^{x} {(1 + logx)}^{2}$

$\rm\implies \:\boxed{ \rm{ \:\rm \: \dfrac{ {d}^{2} y}{d {x}^{2} } = {x}^{x - 1} \bigg(1+ x {(1 + logx)}^{2} \bigg) \: \: }}$

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Formulae Used :-

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

$\boxed{ \rm{ \:\dfrac{d}{dx}logx \: = \: \frac{1}{x} \: }}$

$\boxed{ \rm{ \:\dfrac{d}{dx}k \: = \: 0 \: }}$

$\boxed{ \rm{ \:\dfrac{d}{dx}(u.v) \: = u\dfrac{d}{dx}v \: + \:v\dfrac{d}{dx}u \: \: }}$

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Additional information :-

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

Answer 2

Given function is

$\begin{gathered}\rm \: y = {x}^{x} \\ \end{gathered}$

On taking log on both sides, we get

$\begin{gathered}\rm \: logy = log({x}^{x}) \\ \end{gathered}$

can be rewritten as

$\begin{gathered}\rm \: logy = x \: logx \\ \end{gathered}$

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

$\begin{gathered}\rm \: \dfrac{d}{dx}logy = \dfrac{d}{dx}( x \: logx) \\ \end{gathered}$

$\begin{gathered}\rm \: \dfrac{1}{y}\dfrac{dy}{dx} = x\dfrac{d}{dx}logx + logx\dfrac{d}{dx}x \\ \end{gathered}$

$\begin{gathered}\rm \: \dfrac{1}{y}\dfrac{dy}{dx} = x \times \frac{1}{x} + logx \times 1 \\ \end{gathered}$

$\begin{gathered}\rm \: \dfrac{1}{y}\dfrac{dy}{dx} = 1 + logx \\ \end{gathered}$

$\begin{gathered}\rm\implies \:\dfrac{dy}{dx} = y(1 + logx) \\ \end{gathered}$

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

$\begin{gathered}\rm \: \dfrac{d}{dx}\bigg(\dfrac{dy}{dx}\bigg) = y\dfrac{d}{dx}(1 + logx) + (1 + logx)\dfrac{d}{dx}y \\ \end{gathered}$

$\begin{gathered}\rm \: \dfrac{ {d}^{2} y}{d {x}^{2} } = y \times \dfrac{1}{x} + (1 + logx)\dfrac{dy}{dx} \\ \end{gathered}$

$\begin{gathered}\rm \: \dfrac{ {d}^{2} y}{d {x}^{2} } = \dfrac{ {x}^{x} }{x} + (1 + logx) \times y(1 + logx)\\ \end{gathered}$

$\begin{gathered}\rm \: \dfrac{ {d}^{2} y}{d {x}^{2} } = \dfrac{ {x}^{x} }{x} + {x}^{x} {(1 + logx)}^{2} \\ \end{gathered}$

$\begin{gathered}\rm \: \dfrac{ {d}^{2} y}{d {x}^{2} } = {x}^{x - 1} + {x}^{x} {(1 + logx)}^{2} \\ \end{gathered}$

$\begin{gathered}\rm\implies \:\boxed{ \rm{ \:\rm \: \dfrac{ {d}^{2} y}{d {x}^{2} } = {x}^{x - 1} \bigg(1+ x {(1 + logx)}^{2} \bigg) \: \: }}\\ \end{gathered}$

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