Differentiation Question 24
Question: $ x\sqrt{1+y}+y\sqrt{1+x}=0 $ , then $ \frac{dy}{dx}= $
[RPET 1989, 96]
Options:
A) $ 1+x $
B) $ {{(1+x)}^{-2}} $
C) $ -{{(1+x)}^{-1}} $
D) $ -{{(1+x)}^{-2}} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ x\sqrt{1+y}+y\sqrt{1+x}=0 $
Therefore $ x^{2}(1+y)=y^{2}(1+x) $
Therefore $ {{( \frac{d^{2}s}{dt^{2}} )} _{t=1/2}}=-8 $
Therefore $ x+y+xy=0,{ \because x\ne y } $
Therefore $ \frac{dy}{dx}=\frac{-1}{{{(1+x)}^{2}}} $ .
BETA
