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')