Question

If xsin(a+y)=siny, then \[\frac{dy}{dx}=\]
a)\[\frac{\sin^2 (a+y)}{\cos y}\]
b)\[\frac{\sin a}{\cos^2 (a+y)}\]
c)\[\frac{\sin^2 (a+y)}{\sin a}\]
d)\[\frac{\sin a}{\sin^2 (a+y)}\]

Answer 1

Take the derivative implicitly...
Can you??

Answer 2

shouldn't it be equal to \[\frac{\sin (a+y)}{\cos y-x\cos (a+y)}\]

Answer 3

Firstly you can separate y and x terms:

\[x = \frac{\sin(y)}{\sin(a + y)}\]

Now take the derivative implicitly:

\[dx = \frac{\sin(a+y)(cosy) - \sin(y)(\cos(a + y)}{\sin^2(a+y)}dy\]

\[\frac{dy}{dx} = \frac{\sin^2(a + y)}{\sin(a + y - y)} \implies \frac{dy}{dx} = \frac{\sin^2(a + y)}{\sin(a)}\]

Answer 4

\[\sin(A - B) = sinAcosB - cosAsinB\]

I have used this in the numerator..

Answer 5

I mean numerator after solving will be equal to this..

Answer 6

u could let
\[x=\frac{\sin y}{\sin (a+y)} \]
in what u got up there

Answer 7

yeah, i just did that.. That gives the result.. I don't know why i didn't do that earlier..