Differentiation Question 457
Question: If $ x^{y}=y^{x}, $ then $ \frac{dy}{dx}= $
[DSSE 1996; MP PET 1997]
Options:
A) $ \frac{y(x{\log _{e}}y+y)}{x(y{\log _{e}}x+x)} $
B) $ \frac{y(x{\log _{e}}y-y)}{x(y{\log _{e}}x-x)} $
C) $ \frac{x(x{\log _{e}}y-y)}{y(y{\log _{e}}x-x)} $
D) $ \frac{x(x{\log _{e}}y+y)}{y(y{\log _{e}}x+x)} $
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Answer:
Correct Answer: B
Solution:
$ x^{y}=y^{x}\Rightarrow y{\log _{e}}x=x{\log _{e}}y $
Differentiating w.r.t. x of y, we get $ {\log _{e}}x\frac{dy}{dx}+\frac{y}{x}={\log _{e}}y+x\frac{1}{y}\frac{dy}{dx} $
$ \therefore \frac{dy}{dx}=\frac{y(x{\log _{e}}y-y)}{x(y{\log _{e}}x-x)} $ .
BETA
