If xy=log⁡x then dydx at x=e is

xy=log⁡x⇒log⁡xy=log⁡(log⁡x)⇒ylog⁡x=log⁡(log⁡x)⇒ddx(ylog⁡x)=ddx[log⁡(log⁡x)]y1x+log⁡xdydx=1log⁡x1x⇒log⁡xdydx=1x1log⁡x−ydydx=1xlog⁡x1log⁡x−y=1xlog⁡x1−log⁡(log⁡x)log⁡x At x=e,dydx=1elog⁡e1−log⁡(log⁡e)log⁡e=1e1−log⁡11=1e

If xy=log⁡x then dydx at x=e is

xy=log⁡x⇒log⁡xy=log⁡(log⁡x)⇒ylog⁡x=log⁡(log⁡x)⇒ddx(ylog⁡x)=ddx[log⁡(log⁡x)]y1x+log⁡xdydx=1log⁡x1x⇒log⁡xdydx=1x1log⁡x−ydydx=1xlog⁡x1log⁡x−y=1xlog⁡x1−log⁡(log⁡x)log⁡x At x=e,dydx=1elog⁡e1−log⁡(log⁡e)log⁡e=1e1−log⁡11=1e

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

If $x^{y} = \log x$ then $\frac{dy}{dx}$ at $x = e$ is

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

$e$

$1 / e$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$x^{y} = \log x$

$\begin{array}{l} \Rightarrow \log (x^{y}) = \log (\log x) \Rightarrow ylog x = \log (\log x) \Rightarrow \frac{d}{dx} (ylog x) = \frac{d}{dx} [\log (\log x)] \\ y \frac{1}{x} + \log x \frac{dy}{dx} = \frac{1}{\log x} \frac{1}{x} \Rightarrow \log x \frac{dy}{dx} = \frac{1}{x} [\frac{1}{\log x} - y] \\ \frac{dy}{dx} = \frac{1}{xlog x} [\frac{1}{\log x} - y] = \frac{1}{xlog x} [\frac{1 - \log (\log x)}{\log x}] \\ At x = e, \frac{dy}{dx} = \frac{1}{elog e} [\frac{1 - \log (\log e)}{\log e}] = \frac{1}{e} [\frac{1 - \log 1}{1}] = \frac{1}{e} \end{array}$