how to differentiate x + sin(y)=xy?

Question

I need help with this problem:

x + sin(y) = xy

what does dy/dx = ?

Answer 1

Implicit differentiation is the easier way to go here and it's probably the way they wanted it done. I'll do it a second way lower down and you can compare.
When doing implicit diff you have to remember y is some function of x so its deriv is given by y' and the derivative of xy is the deriv of a product.
d(xy)/dx = xy' + y. Make sure this is okay with you.
dsiny/dx = cosy*y' which is like dsinf(x)/dx =cosf(x)*f'(x).
Putting it all together:
1 +(cosy) y' = xy' + y; Now solve for y' and you have dydx.
The other way;
solve for x in terms of y:
x = siny/(y - 1) Use the quotient rule to find dx/dy =
[(y - 1)cosy - siny]/(y - 1)^2
Now take the reciprocal and you have dy/dx

Answer 2

Actually, Will, sin'=cos, not -cos.

Anyhow, let's write sin(y) = y (x-1), and then

sin(y)/y = x-1.

Differentiate both sides with respect to x.

If I'm recalling the Quotient Rule correction, we have

(y(sin(y))' - y'sin(y))/y^2 = 1

y' * (ycosy - siny)/y^2 = 1.

Hmm. We know how to express siny/y as a function of x, and hence cosy/y also, but I'm drawing a blank at the moment on how to express siny/y^2 in terms of x ...

Answer 3

subtract x from both sides
sin(y)=xy-x

simplify
sin(y) = x(y-1)

the derivative of sin is negative cosine I think, so
dy/dx (sin(y))= dy/dx (x(y-1))

-cos(y) = (y-1)
or
cos(y) = 1-y