Question

Using the function

${x}^{ \frac{1}{x} } \: and \: x > 0 \: prove \: that \: {e}^{\pi} > {\pi}^{e}$
​

Answer 1

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

Given function is

$\rm :\longmapsto\:f(x) = {\bigg(x\bigg) }^{\dfrac{1}{x} }$

On taking log on both sides, we get

$\rm :\longmapsto\:logf(x) = log {\bigg(x\bigg) }^{\dfrac{1}{x} }$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ log {x}^{y} = y \: logx \: }}}$

So, using this, we get

$\rm :\longmapsto\:logf(x) = \dfrac{1}{x} \: logx$

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

$\rm :\longmapsto\:\dfrac{d}{dx}logf(x) = \dfrac{d}{dx}\bigg(\dfrac{1}{x} \: logx\bigg)$

$\rm :\longmapsto\:\dfrac{1}{f(x)}f'(x) = \dfrac{1}{x}\dfrac{d}{dx}logx + logx\dfrac{d}{dx} \dfrac{1}{x}$

We know,

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

and

$\purple{\rm :\longmapsto\:\boxed{\tt{ \dfrac{d}{dx} \frac{1}{ {x}^{n} } \: = \: \frac{ - n}{ {x}^{n + 1} } \: }}}$

So, using this result, we get

$\rm :\longmapsto\:f'(x) = f(x)\bigg[\dfrac{1}{x} \times \dfrac{1}{x} + logx\bigg(\dfrac{ - 1}{ {x}^{2} } \bigg) \bigg]$

$\rm :\longmapsto\:f'(x) = f(x)\bigg[\dfrac{1}{ {x}^{2} } + logx\bigg(\dfrac{ - 1}{ {x}^{2} } \bigg) \bigg]$

$\rm :\longmapsto\:f'(x) = \dfrac{f(x)}{ {x}^{2} } \bigg[1 - logx\bigg]$

Now, For critical points, we have f'(x) = 0

$\rm :\longmapsto\:1 - logx = 0$

$\rm :\longmapsto\:- logx = - 1$

$\rm :\longmapsto\:logx = 1$

$\rm :\longmapsto\:logx = loge$

$\bf\implies \:x = e$

Now, when

$\rm :\longmapsto\:x > e$

$\rm :\longmapsto\:f'(x) < 0$

$\bf\implies \:f(x) \: is \: decreasing \: function$

So, By using Definition of decreasing functions,

$\rm :\longmapsto\:\pi > e \: \: \rm\implies \:f(\pi) < f(e)$

$\rm :\longmapsto\:{\bigg(\pi\bigg) }^{\dfrac{1}{\pi} } < {\bigg(e\bigg) }^{\dfrac{1}{e} }$

$\rm\implies \: {\pi}^{e} < {e}^{\pi}$

$\bf\implies \: {e}^{\pi} > {\pi}^{e}$

Hence, Proved

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FORMULA USED

$\boxed{\tt{ \dfrac{d}{dx} log(x) = \frac{1}{x} \: }}$

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

$\boxed{\tt{ log(e) = 1 \: }}$

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LEARN MORE

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

Answer 2

Answer:

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

Given function is

$\rm :\longmapsto\:f(x) = {\bigg(x\bigg) }^{\dfrac{1}{x} }$

On taking log on both sides, we get

$\rm :\longmapsto\:logf(x) = log {\bigg(x\bigg) }^{\dfrac{1}{x} }$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ log {x}^{y} = y \: logx \: }}}$

So, using this, we get

$\rm :\longmapsto\:logf(x) = \dfrac{1}{x} \: logx$

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

$\rm :\longmapsto\:\dfrac{d}{dx}logf(x) = \dfrac{d}{dx}\bigg(\dfrac{1}{x} \: logx\bigg)$

$\rm :\longmapsto\:\dfrac{1}{f(x)}f'(x) = \dfrac{1}{x}\dfrac{d}{dx}logx + logx\dfrac{d}{dx} \dfrac{1}{x}$

We know,

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

and

$\purple{\rm :\longmapsto\:\boxed{\tt{ \dfrac{d}{dx} \frac{1}{ {x}^{n} } \: = \: \frac{ - n}{ {x}^{n + 1} } \: }}}$

So, using this result, we get

$\rm :\longmapsto\:f'(x) = f(x)\bigg[\dfrac{1}{x} \times \dfrac{1}{x} + logx\bigg(\dfrac{ - 1}{ {x}^{2} } \bigg) \bigg]$

$\rm :\longmapsto\:f'(x) = f(x)\bigg[\dfrac{1}{ {x}^{2} } + logx\bigg(\dfrac{ - 1}{ {x}^{2} } \bigg) \bigg]$

$\rm :\longmapsto\:f'(x) = \dfrac{f(x)}{ {x}^{2} } \bigg[1 - logx\bigg]$

Now, For critical points, we have f'(x) = 0

$\rm :\longmapsto\:1 - logx = 0$

$\rm :\longmapsto\:- logx = - 1$

$\rm :\longmapsto\:logx = 1$

$\rm :\longmapsto\:logx = loge$

$\bf\implies \:x = e$

Now, when

$\rm :\longmapsto\:x > e$

$\rm :\longmapsto\:f'(x) < 0$

$\bf\implies \:f(x) \: is \: decreasing \: function$

So, By using Definition of decreasing functions,

$\rm :\longmapsto\:\pi > e \: \: \rm\implies \:f(\pi) < f(e)$

$\rm :\longmapsto\:{\bigg(\pi\bigg) }^{\dfrac{1}{\pi} } < {\bigg(e\bigg) }^{\dfrac{1}{e} }$

$\rm\implies \: {\pi}^{e} < {e}^{\pi}$

$\bf\implies \: {e}^{\pi} > {\pi}^{e}$

Hence, Proved

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

FORMULA USED

$\boxed{\tt{ \dfrac{d}{dx} log(x) = \frac{1}{x} \: }}$

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

$\boxed{\tt{ log(e) = 1 \: }}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

LEARN MORE

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