Differentiation Question 434
Question: If $ x^{y}={e^{x-y}} $ , then $ \frac{dy}{dx}= $
[MP PET 1987, 2004; MNR 1984; Roorkee 1954; BIT Ranchi 1991; RPET 2000]
Options:
A) $ \log x.{{[\log (ex)]}^{-2}} $
B) $ \log x.{{[\log (ex)]}^{2}} $
C) $ \log x.{{(\log x)}^{2}} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ x^{y}={e^{x-y}} $
Therefore $ y\log x=x-y $
Therefore $ y=\frac{x}{1+\log x} $
Therefore $ \frac{dy}{dx}=\log x{{(1+\log x)}^{-2}}=\log x{{[\log ex]}^{-2}} $ .
BETA
