Understanding Implicit Solutions: Solving a Basic Problem with x*sin(y) = cos(y)

[logan3]

hi,

I could you some help with a basic problem please.

it asks to verify that x*sin(y) = cos(y) is an implicit solution to \frac{dy}{dx} (x*cot(y) +1) = -1

Here's what I have so far:
x*sin(y) = cos (y)
\frac{dx}{dx} * sin(y) * ((\frac{d}{dx}) * sin(y)) * (\frac{dy}{dx}) = (\frac{d}{dx}) cos(y) * (\frac{dy}{dx})
\rightarrow sin(y) * cos(y) * \frac{dy}{dx} = -sin(y) * \frac{dy}{dx}

I'm stuck here. For example, if I try to divide both sides by sin(y) in order to elimnate sin(y), then I can't figure out how to get to the d.e. from there:
ex. (sin(y) * cos(y) * \frac{dy}{dx}) / sin(y) = -(sin(y) * \frac{dy}{dx}) / sin(y)
\rightarrow cos(y) * \frac{dy}{dx} = - \frac{dy}{dx}
add \frac{dy}{dx} to both sides
\rightarrow cos(y) * \frac{dy}{dx} + \frac{dy}{dx} = 0
factor
\rightarrow \frac{dy}{dx} * (cos(y) + 1) = 0

thanks for any help.

 

[Office_Shredder]

Doing the product rule on x*sin(y) should involve adding some functions together at the end

 

[logan3]

thanks, I got the answer after adding in the product rule:

x*sin(y) = cos(y)

take the derivative of both sides

\rightarrow \frac{dx}{dx}*sin(y) + \frac{dy}{dx}*cos(y)*x = \frac{dy}{dx}*(-sin(y))

divide both sides by sin(y)

\rightarrow 1 + \frac{dy}{dx}*\frac{cos(y)}{sin(y)}*x = -\frac{dy}{dx}

subtract -1 from both sides and add \frac{dy}{dx} from both sides

\rightarrow \frac{dy}{dx}*cot(y)*x + \frac{dy}{dx} = -1

factor

\rightarrow \frac{dy}{dx}*(x*cot(y) + 1) = -1