Question

The function $f\left( x \right)={{x}^{\ln \left( \ln x \right)}}$ has,
(a) local minimum at $x=\sqrt[e]{e}$
(b) local maximum at $x={{e}^{e}}$
(c) local minimum at $x=\sqrt[e+1]{e}$
(d) local maximum at $x=\dfrac{1}{{{e}^{e}}}$

Answer 1

Write the given function in terms of exponential function by first taking log function both sides and then again converting back in f (x) by power of ‘e’ conversion formula. Now, differentiate f (x) and substitute it equal to zero. Find the values of x. Again, differentiate f’ (x) to find f’’ (x) and substitute the value of x found earlier in f’’ (x). If f’’ (x) is positive then the value of ‘x’ will be a point of minima and if f’’ (x) is negative then the value of ‘x’ will be a point of maxima.

Complete answer:
We have been provided with the function, $f\left( x \right)={{x}^{\ln \left( \ln x \right)}}$.
Taking natural log ‘ln’ both sides, we get,
$\Rightarrow \ln f\left( x \right)=\ln \left[ {{x}^{\ln \left( \ln x \right)}} \right]$
Using the property: - $\ln {{x}^{a}}=a\ln x$, we get,
$\Rightarrow \ln f\left( x \right)=\ln \left( \ln x \right)\times \ln x$
This can be written as,
$\Rightarrow f\left( x \right)={{e}^{\ln \left( \ln x \right)\times \ln x}}$
Differentiating the above function using chain rule, we get,

& \Rightarrow f'\left( x \right)=\dfrac{d}{dx}\left[ {{e}^{\ln \left( \ln x \right)\times \ln x}} \right] \\\ & \Rightarrow f'\left( x \right)=\dfrac{d\left[ {{e}^{\ln \left( \ln x \right)\times \ln x}} \right]}{d\left[ \ln \left( \ln x \right)\times \ln x \right]}\times \dfrac{d\left[ \ln \left( \ln x \right)\times \ln x \right]}{dx} \\\ & \Rightarrow f'\left( x \right)=\left( {{e}^{\ln \left( \ln x \right)\times \ln x}} \right)\times \left[ \ln x.\dfrac{d\ln \left( \ln x \right)}{dx}+\ln \left( \ln x \right).\dfrac{d\left[ \ln x \right]}{dx} \right] \\\ & \Rightarrow f'\left( x \right)=f\left( x \right)\times \left[ \ln x\times \dfrac{1}{\ln x}\times \dfrac{1}{x}+\ln \left( \ln x \right)\times \dfrac{1}{x} \right] \\\ \end{aligned}$$ $$\Rightarrow f'\left( x \right)=f\left( x \right)\times \left[ \dfrac{1+\ln \left( \ln x \right)}{x} \right]$$ - (1) Substituting f’ (x) = 0, we get, $$\begin{aligned} & \Rightarrow f\left( x \right)\times \left[ \dfrac{1+\ln \left( \ln x \right)}{x} \right]=0 \\\ & \Rightarrow f\left( x \right)\times \left[ 1+\ln \left( \ln x \right) \right]=0 \\\ \end{aligned}$$ Since, f (x) is an exponential function, therefore it cannot be 0 or negative. Therefore, $$\begin{aligned} & \Rightarrow 1+\ln \left( \ln x \right)=0 \\\ & \Rightarrow \ln \left( \ln x \right)=-1 \\\ & \Rightarrow \ln x={{e}^{-1}} \\\ & \Rightarrow \ln x=\dfrac{1}{e} \\\ & \Rightarrow x={{e}^{\dfrac{1}{e}}} \\\ \end{aligned}$$ So, we have only one value of x for which f’ (x) = 0. So, let us check if this x is a point of maxima or minima. Now, differentiating f’ (x), we get, $$\begin{aligned} & \Rightarrow f''\left( x \right)=f\left( x \right)\times d\left[ \dfrac{1+\ln \left( \ln x \right)}{x} \right]+\left[ \dfrac{1+\ln \left( \ln x \right)}{x} \right]\times \dfrac{d\left[ f\left( x \right) \right]}{dx} \\\ & \Rightarrow f''\left( x \right)=f\left( x \right)\times \left[ \dfrac{\left( x\times \dfrac{1}{\ln x}\times \dfrac{1}{x} \right)-\left[ \left( 1+\ln \left( \ln x \right) \right) \right]\times 1}{{{x}^{2}}} \right]+\left[ \dfrac{1+\ln \left( \ln x \right)}{x} \right]\times f'\left( x \right) \\\ & \Rightarrow f''\left( x \right)=\dfrac{f\left( x \right)}{{{x}^{2}}}\left[ \dfrac{1}{\ln x}-\left( 1+\ln \left( \ln x \right) \right) \right]+\left[ \dfrac{1+\ln \left( \ln x \right)}{x} \right]\times f'\left( x \right) \\\ \end{aligned}$$ Using equation (1), we can write f’’ (x) as: - $$\begin{aligned} & \Rightarrow f''\left( x \right)=\dfrac{f\left( x \right)}{{{x}^{2}}}{{\left[ \dfrac{1}{\ln x}-\left( 1+\ln \left( \ln x \right) \right)+\left[ \dfrac{\left[ 1+\ln \left( \ln x \right) \right]}{x} \right] \right]}^{2}}f\left( x \right) \\\ & \Rightarrow f''\left( x \right)=\dfrac{f\left( x \right)}{{{x}^{2}}}\left[ \dfrac{1}{\ln x}-\left( 1+\ln \left( \ln x \right) \right)+{{\left[ 1+\ln \left( \ln x \right) \right]}^{2}} \right] \\\ \end{aligned}$$ Now, at $$x={{e}^{\dfrac{1}{e}}}$$, we have, $$\begin{aligned} & \Rightarrow f''\left( x \right)=f''\left( {{e}^{\dfrac{1}{e}}} \right)=\dfrac{f\left( {{e}^{\dfrac{1}{e}}} \right)}{{{\left( {{e}^{\dfrac{1}{e}}} \right)}^{2}}}\left[ \dfrac{1}{\ln {{e}^{\dfrac{1}{e}}}}-\left( 1+\ln \left( \ln {{e}^{\dfrac{1}{e}}} \right) \right)+{{\left[ 1+\ln \left( \ln {{e}^{\dfrac{1}{e}}} \right) \right]}^{2}} \right] \\\ & \Rightarrow f''\left( x \right)=f''\left( {{e}^{\dfrac{1}{e}}} \right)=\dfrac{f\left( {{e}^{\dfrac{1}{e}}} \right)}{{{e}^{\dfrac{2}{e}}}}\left[ \dfrac{1}{\dfrac{1}{e}}-\left( 1+\ln \dfrac{1}{e} \right)+{{\left[ 1+\ln \dfrac{1}{e} \right]}^{2}} \right] \\\ & \Rightarrow f''\left( x \right)=f''\left( {{e}^{\dfrac{1}{e}}} \right)=\dfrac{f\left( {{e}^{\dfrac{1}{e}}} \right)}{{{e}^{\dfrac{2}{e}}}}\left[ e-\left( 1+\left( -1 \right) \right)+{{\left( 1+\left( -1 \right) \right)}^{2}} \right] \\\ & \Rightarrow f''\left( x \right)=f''\left( {{e}^{\dfrac{1}{e}}} \right)=\dfrac{f\left( {{e}^{\dfrac{1}{e}}} \right)}{{{e}^{\dfrac{2}{e}}}}\times e \\\ \end{aligned}$$ Since, all the terms in the above expression are exponential functions, therefor they cannot be negative. Hence, f’’ (x) at $$x={{e}^{\dfrac{1}{e}}}$$ will always be positive. $$\Rightarrow f''\left( {{e}^{\dfrac{1}{e}}} \right)>0$$ Therefore, $$x={{e}^{\dfrac{1}{e}}}$$ is a point of minima. Now, $$x={{e}^{\dfrac{1}{e}}}$$ can be written as $$x=\sqrt[e]{e}$$. **Hence, option (a) is the correct answer.** **Note:** One may note that we can apply the first derivative test, instead of the second derivative test to find the correct option. In the first derivative test we will substitute f’ (x) = 0 to find the value of x. Now, we will put this value of x on the number line, if sign of f (x) changes from negative to positive then ‘x’ will be a point of minima and if sign of f (x) changes from positive to negative then ‘x’ will be a point of maxima.