How do I solve for y' in implicit differentiation problems?

[phantomcow2]

Homework Statement

Find derivative of y with respect to x.

Sin(xy) = Sinx Siny

Homework Equations

The Attempt at a Solution

Use chain rule (product rule for inner function) to differentiate the left. Use product rule to differentiate the right and I get the following:

cos(xy)(y+xy') = (cosx siny) + (cosy' sinx)

distribute the cos(xy) on the left to get:
ycos(xy) + xy'cos(xy) = (cosx siny) + (cosy' sinx)

Rearrange to get y' on one side, everything else on the other.

ycos(xy) - cosx siny = -xy'cos(xy) + cosy' sinx

Now what? I don't understand how I'd solve for y' from here. Inverse cosine?

 

[quantumdude]

cos(xy)(y+xy') = (cosx siny) + (cosy' sinx)

OOOOHHH, so close! You got the left side correct, and you got the first term on the right side correct. But you got the second term wrong. When you take the derivative of $\cos(y)$ with respect to $x$ you have to use the chain rule.

 

[Mark44]

You have an error here:

cos(xy)(y+xy') = (cosx siny) + (cosy' sinx)

d/dx(cos(y)) isn't cos(y')