Question

If y=xxx....[infinity], then prove that dydx=y2x(1−ylogx)

Answer 1

Question :-

$\sf{if \: \: \: y \: = {x}^{ {x}^{ {x}^{...... \infty} } } }$

$\bf{then \: prove \: that} \: \\ \sf{ \frac{dy}{dx} \: = \frac{ {y}^{2}}{ x \big(1 - y log(x) \big)}}$

Formula used :-

$\rightarrow \: \boxed{\sf{ \frac{d}{dx} u \: v \: = u \: \frac{d}{dx} v \: + v \: \frac{d}{dx} u}} \\ \\ \rightarrow \: \boxed{ \sf{log \: {m}^{n} = n \: log \: m}}$

Explanation :-

According to the question,

$\rightarrow \: \sf{y \: = {x}^{ {x}^{ {x}^{...... \infty} } } } \\ \\ \: \bf{ we \: can \: write \: it \: } \\ \\ \rightarrow \: \sf{y \: = {x}^{y} } \\ \\ \sf{taking \: log \: on \: both \: sides \: } \\ \\ \sf{ \rightarrow \: log \: y \: = log \: {x}^{y} } \\ \\ \because \: \boxed{ \sf{log \: {m}^{n} = n \: log \: m}}\\ \\ \rightarrow \: \sf{log \: y \: = y \: log \: x}$

Now differentiate with respect to "x",

$\rightarrow \: \frac{1}{y} \sf{ \frac{dy}{dx} = y \frac{d}{dx} log \: x + log \: x \: \frac{d}{dx} y} \\ \\ \rightarrow \: \sf{ \frac{1}{y} \frac{dy}{dx} = \frac{y}{x} + log \: x \: \frac{dy}{dx} } \\ \\ \rightarrow \: \sf{ \frac{1}{y} \frac{dy}{dx} \: - log \: x \: \frac{dy}{dx} \: = \frac{y}{x} } \\ \\ \rightarrow \: \sf{ \frac{dy}{dx} \bigg( \frac{1}{y} - log \: x \bigg) = \frac{y}{x} } \\ \\ \rightarrow \: \sf{\frac{dy}{dx} \bigg( \frac{1 - y \: log \: x}{y} \bigg) = \frac{y}{x} } \\ \\ \rightarrow \: \boxed{ \sf{ \frac{dy}{dx} \: = \frac{ {y}^{2} }{x(1 - \: y \: log \: x)} }}$

Hence proved .

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