Question
If $$f\left( x \right) = {\log _x}\left( {\ln x} \right),$$ then at $$x = e,\,f'\left( x \right)$$ equals :
A.
$$0$$
B.
$$1$$
C.
$$e$$
D.
$$\frac{1}{e}$$
Answer :
$$\frac{1}{e}$$
Solution :
$$\eqalign{
& \because \,\ln \,x = {\log _e}x,\,\,{\text{so}} \cr
& f\left( x \right) = {\log _x}\left( {{{\log }_e}x} \right) = \frac{{\log \left( {\log \,x} \right)}}{{\log \,x}} \cr
& \Rightarrow f'\left( x \right) = \frac{{\log \,x\left( {\frac{1}{{x\,\log \,x}}} \right) - \log \left( {\log \,x} \right).\frac{1}{x}}}{{{{\left( {\log \,x} \right)}^2}}} \cr
& \therefore \,f'\left( e \right) = \frac{{\frac{1}{e} - 0}}{{{{\left( 1 \right)}^2}}} = \frac{1}{e} \cr} $$