Differentiation Question 57
Question: If $ \sin y=x\sin (a+y), $ then $ \frac{dy}{dx}= $
[Karnataka CET 2000; UPSEAT 2001; Pb. CET 2001; Kerala (Engg.) 2005]
Options:
A) $ \frac{{{\sin }^{2}}(a+y)}{\sin (a+2y)} $
B) $ \frac{{{\sin }^{2}}(a+y)}{\sin (a+2y)} $
C) $ \frac{{{\sin }^{2}}(a+y)}{\sin a} $
D) $ \frac{{{\sin }^{2}}(a+y)}{\cos a} $
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Answer:
Correct Answer: C
Solution:
$ \sin y=x\sin (a+y) $
Therefore $ x=\frac{\sin y}{\sin (a+y)} $
Therefore $ 1=\frac{\cos y.\frac{dy}{dx}.\sin (a+y)-\sin y\cos (a+y)\frac{dy}{dx}}{{{\sin }^{2}}(a+y)} $
$ =\frac{\frac{dy}{dx}.\sin (a+y-y)}{{{\sin }^{2}}(a+y)}\Rightarrow \frac{dy}{dx}=\frac{{{\sin }^{2}}(a+y)}{\sin a} $ .
BETA
