Question

Differentiate x^x^x w.r.tx

Answer 1

Correct Question:

Differentiate

${ {x}^{x} }^{x}$

W.r.t.x

Solution:

$let \: y = { {x}^{x} }^{x} .......(1)$

Taking log on both side,

$log \: y = log { {x}^{x} }^{x} = {x}^{x} logx...........(2)$

Again taking log on both sides , we get;

$log(log \: y) = log( {x}^{x} \: logx) \\ \\ = log {x}^{x} + log(logx) \\ \\ [log \: mn = log \: m + log \: n] \\ \\ = x \: logx + log(logx)$

Now differentiating both side w.r.t.x we get,

$\frac{1}{log \: y} . \frac{d}{dx} (log \: y) = [x \: \frac{d}{dx} (log \: x) + log \: x. \frac{d}{dx} (x)] + \frac{1}{logx} . \frac{d}{dx} (logx)$

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By Product Rule,

$\frac{d}{dx}(xy) = x \frac{d}{dx}(y) + y \frac{d}{dx} x$

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$= > \frac{1}{log \: y} . \frac{1}{y} . \frac{dy}{dx} = [x. \frac{1}{x} + log \: x.1] + \frac{1}{log \: x} . \frac{1}{x} \\ \\ = > \frac{1}{y \: log \: y} . \frac{dy}{dx} = (1 + logx) + \frac{1}{xlogx} \\ \\ = > \frac{dy}{dx} = y \: log \: y[1 + log \: x + \frac{1}{x \: log \: x} ]$

Hence, Using (1) and (2) we get

$\frac{dy}{dx} = { {x}^{x} }^{x} \: log \: x[1 + log \: x + \frac{1}{x \: log \: x} ]$