Differentiation Question 124
Question: If then $ \frac{dy}{dx}= $
[DSSE 1980; CBSE 1980]
Options:
A) $ \frac{y[2xy-y^{2}\cos (xy)-1]}{xy^{2}\cos (xy)+y^{2}-x} $
B) $ \frac{[2xy-y^{2}\cos (xy)-1]}{xy^{2}\cos (xy)+y^{2}-x} $
C) $ -\frac{y[2xy-y^{2}\cos (xy)-1]}{xy^{2}\cos (xy)+y^{2}-x} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ \sin (xy)+\frac{x}{y}=x^{2}-y $
Differentiating both sides, $ \cos (xy)\frac{d}{dx}(xy)+x{ -\frac{1}{y^{2}} }\frac{dy}{dx}+\frac{1}{y}=2x-\frac{dy}{dx} $
Therefore $ [x\cos (xy)-\frac{x}{y^{2}}+1]\frac{dy}{dx}=2x-\frac{1}{y}-y\cos (xy) $
Therefore $ \frac{dy}{dx}=[ \frac{2xy^{2}-y-y^{3}\cos (xy)}{xy^{2}\cos (xy)-x+y^{2}} ] $ .
BETA
