If $y = x^{{(\log x)}^{\log (\log x)}}$ then which is/are correct ?

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{d y}{d x} = \frac{y}{x} ((\ln x^{x - 1}) + 2 \ln x \ln (\ln x))$

$\frac{d y}{d x} = \frac{y}{x} {(\log x)}^{\log (\log x)} (2 \log (\log x) + 1)$

$\frac{d y}{d x} = \frac{y}{x \ln x} [{(\ln x)}^{2} + 2 \ln (\ln x)]$

$\frac{d y}{d x} = \frac{y}{x} \frac{\log y}{\log x} [2 \log (\log x) + 1]$

answer is B, D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$y = x^{{(\log x)}^{\log (\log x)}}$
$∴ \log y = (\log x) {(\log x)}^{\log (\log x)}$
$y = x^{{(\log x)}^{\log (\log x)}}$
$∴ \log y = (\log x) {(\log x)}^{\log (\log x)} (1)$
Taking log of both sides, we get
$\log (\log y) = \log (\log x) + \log (\log x) \log (\log x)$
Differentiating w.r.t. $x$ we get
$\frac{1}{\log y} . \frac{1}{y} \frac{d y}{d x} = \frac{1}{x \log x} + \frac{2 \log (\log x)}{\log x} \frac{1}{x}$
$= \frac{2 \log (\log x) + 1}{x \log x}$ or $\frac{d y}{d x} = \frac{y}{x} . \frac{\log y}{\log x} (2 \log (\log x) + 1)$
Substituting the value of $\log y$ from (1), we get
$\frac{d y}{d x} = \frac{y}{x} {(\log x)}^{\log (\log x)} (2 \log (\log x) + 1)$