Question

Please solve the given question

$y = {(cotx)}^{ {(cotx)}^{ {(cotx)}^{ - - - -} } } \: \\ \\ find \: \frac{dy}{dx} \: at \: x = \frac{\pi}{4}$

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Answer 1

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

Given that,

$\rm \: y = {(cotx)}^{ {(cotx)}^{(cotx) - - - } }$

can be rewritten as

$\rm \: y = {(cotx)}^{y}$

On taking log on both sides, we get

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

We know,

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

So, on using this result, we get

$\rm \: logy = y log{(cotx)}$

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

$\rm \: \dfrac{d}{dx}logy \: = \: \dfrac{d}{dx} \: y \: log(cotx)$

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

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

$\rm \: \dfrac{1}{y}\dfrac{dy}{dx} = y \: \times \dfrac{1}{cotx} \times \dfrac{d}{dx}(cotx) + log(cotx)\dfrac{dy}{dx}$

$\rm \: \dfrac{1}{y}\dfrac{dy}{dx} - log(cotx)\dfrac{dy}{dx} = \dfrac{y}{cotx}( - {cosec}^{2}x)$

$\rm \: \bigg(\dfrac{1}{y} - log(cotx)\bigg)\dfrac{dy}{dx} = - \dfrac{y {cosec}^{2} x}{cotx}$

$\rm \: \bigg(\dfrac{1 - ylog(cotx)}{y} \bigg)\dfrac{dy}{dx} = - \dfrac{y {cosec}^{2} x}{cotx}$

$\rm\implies \:\rm \: \dfrac{dy}{dx} = - \dfrac{ {y}^{2} {cosec}^{2} x}{cotx \: [1 - y \: log(cotx)]}$

Now, we have

$\rm \: y = {(cotx)}^{y}$

So,

$\rm \: At \: x = \dfrac{\pi}{4}$

$\rm \: y = {\bigg(cot\dfrac{\pi}{4} \bigg) }^{y}$

$\rm \: y = {1}^{y}$

$\rm\implies \:y = 1$

So, on substituting these values, we get

$\rm \: \dfrac{dy}{dx} = - \: \dfrac{ {1}^{2} {cosec}^{2} \dfrac{\pi}{4}}{cot\dfrac{\pi}{4} \: \bigg[1 - 1 \times \: logcot\dfrac{\pi}{4}\bigg]}$

$\rm \: \dfrac{dy}{dx} = - \: \dfrac{ {( \sqrt{2}) }^{2} }{1 \: \bigg[1 -\: log1\bigg]}$

$\rm \: \dfrac{dy}{dx} = - \: \dfrac{ 2 }{1 \: \bigg[1 -\: 0\bigg]}$

$\rm\implies \:\dfrac{dy}{dx} \: = \: - \: 2$

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Additional Information :-

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

Answer 2

Step-by-step explanation:

$\pink{ \text{Given, \( \tt y=(\cot x)^{(\cot x)^{(\cot x) \cdots \infty}} \)}}$

$\color{red} \text{To find, prove \( \tt \dfrac{d y}{d x}=-2 \) at \( \tt x=\dfrac{\pi}{4} \)}$

$\color{blue} \text{ \boxed{ \boxed{ \tt \: Solution:} } \qquad \: Here, \( \tt y=(\cot x)^{(\cot x)^{\cot x \ldots \infty}} \)}$

$\color{darkcyan} \tt\[ y=(\cot x)^{y} \]$

$\color{darkred} \text{ taking \( \log \) on both sides}$

$\color{violet} \: \: \tt\[ \log y=\log (\cot x)^{y} \] \\ \color{violet}\tt \[ \log y=y \log \cot x \]$

$\text{differentiating with respect to \( \sf x .\)}$

$\color{darkviolet}\[ \begin{aligned} \tt\frac{1}{y} \frac{d y}{d x} & \tt=y \frac{d}{d x} \log \cot x+\log \cot \frac{d y}{d x} \\ \\ \tt \frac{1}{y} \frac{d y}{d x} & \tt=\frac{y}{\cot x} \frac{d}{d x} \cot x+\log \cot x \frac{d y}{d x} \\ \\ \tt\frac{d y}{d x}\left[\frac{1}{y}-\log \cot x\right] &\tt=\frac{y}{\cot x} \operatorname{cosec}^{2} x \\ \\ \tt \left[\frac{d y}{d x}\right]_{x=\frac{\pi}{4}} & \tt\frac{y \operatorname{cosec}^{2} \frac{\pi}{4}}{\cot \frac{\pi}{4}} \cdot \tt\frac{y}{1-y \log \cot \frac{\pi}{4}} \\ \\ \tt\left[\frac{d y}{d x}\right]_{\frac{\pi}{4}} &\tt=\frac{y^{2}(\sqrt{2})^{2}}{1(1-y \log \cot 1]} \\ \\ & \tt= \frac{2( - 1) {}^{2} }{ - 1 - 0} \\ \\ \tt\left[\frac{d y}{d x}\right]_{\frac{\pi}{4}} &\tt=-2 \\ \\ &\text{Hence proved. }\end{aligned} \]$