Question

$\colorbox{red} {\boxed{ \color{white}\text{if \: } \tt \: y=x^{x^{x}} \: then \: \: \: find \: \: \dfrac{dy}{dx} .}}$
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Answer 1

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

Given function is

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

On taking log on both sides, we get

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

We know,

$\boxed{\sf{ \:log {x}^{y} = ylogx \: }}$

So, using this result, we get

$\rm \: logy = {x}^{x} \: logx$

Again, taking log on both sides, we get

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

We know,

$\boxed{\sf{ \:logxy = logx + logy \: }}$

So, using this result, we get

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

can be further rewritten as

$\rm \: log(logy) = xlogx + log(logx)$

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

$\rm \: \dfrac{d}{dx}log(logy) =\dfrac{d}{dx}\bigg( xlogx + log(logx) \bigg)$

We know,

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

and

$\boxed{\sf{ \:\dfrac{d}{dx}uv \: = \: u\dfrac{d}{dx}v + \: v\dfrac{d}{dx}u \: }}$

So, using these results, we get

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

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

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

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

On substituting the values of y and logy from above, we get

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

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

Hence,

$\rm\implies \:\boxed{\sf{ \:\rm \: \dfrac{dy}{dx} = {x}^{ {x}^{x} } \: {x}^{x} \: \bigg(logx + (logx)^{2} + \dfrac{1}{x}\bigg) \: }}$

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Additional Information :-

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