Calculus problem #2152

$y = {(\ln{x})}^{x}$ $=>\ln{y} = x\ln{(\ln{x})}$ $=>\frac{1}{y}\frac{dy}{dx} = x\times{\frac{1}{\ln{x}}}\times{\frac{1}{x}}+\ln{(\ln{x})}$ $=>\frac{dy}{dx} = y\times(x\times{\frac{1}{\ln{x}}}\times{\frac{1}{x}}+\ln{(\ln{x})})$ $=>\frac{dy}{dx} = {(\ln{x})}^{x}\times(x\times{\frac{1}{\ln{x}}}\times{\frac{1}{x}}+\ln{(\ln{x})})$

Putting x = e derivative is: ${(\ln{e})}^{e}\times(e\times{\frac{1}{\ln{e}}}\times{\frac{1}{e}}+\ln{(\ln{e})})$ $=\boxed{1}$

At $x = e$ , $y = (\ln e)^{e} = 1$

$y = (\ln x)^{x}$

$\ln y = x \times \ln(\ln x)$

Differentiating with respect to x,

$\frac{1}{y} \times \frac{dy}{dx} = \ln(\ln x) + \frac{1}{\ln x }$

At x = e and y = 1,

$1 \times \frac{dy}{dx} = 0 + 1 = 1$

taking log both sides and then differentiating so that dy/dx=y.{ln(lnx)+1/lnx}

y=(ln x)^x taking log base 'e' on both the sides we arrive at, =1/y* dy/dx=ln( ln x) + x (1/ln(x) ) * (1/x) {Using chain rule} =1 dy/dx= 0+1 {at x=e} =1

$y = (\ln x)^x$

$\ln y = x . \ln(\ln x))$

$\frac{y'}{y} = 1 . \ln(ln x) + \frac{1}{\ln x} . \frac{1}{x} . x$

$y' = y . (\ln(\ln x) + \frac{1}{\ln x}) .$

$y' = (\ln x)^x . (\ln(\ln x) + \frac{1}{\ln x})$

When $x = e$ , then

$y'(e) = (\ln e)^e . (\ln(\ln e) + \frac{1}{\ln e})$

$y'(e) = 1^e . (\ln 1 + \frac{1}{1})$

$y'(e) = 1 . (0 + 1)$

$\boxed{ y'(e) = 1 }$

Actually, all you have to do is to find the derivative of { (\ln { x) } }^{ x } and then (using power rule and derivative of natural log) you get \frac { 1 }{ x } x, which is just 1 (the answer!)

y=({ \ln { x } ) }^{ x }\quad \ y=({ \ln { e } })^{ e }\quad where\quad x=e\ y={ 1 }^{ e }\ y=1