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

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