using implicit differentiation, find y' for sin(x) + sin(x)sin(y) = cos(xy)

Question

I tried it multiple times but couldn't get the right answer, so could someone show me the specific steps for this question?

Thanks!

Answer 1

By the chain rule,

d/dx[sin y] = (cos y) y'

so that, by the product rule,

d/dx[sin x sin y] = cos x sin y + sin x (cos y) y'

Therefore,

d/dx[sin x + sin x sin y] = cos x + cos x sin y + sin x (cos y) y' = cos x (1 + sin y) + sin x (cos y) y'

The chain and product rules yield

d/dx[cos(xy)] = -sin(xy) d/dx(xy) = -sin(xy) (y + xy')

Hence

cos x (1 + sin y) + sin x (cos y) y' = -sin(xy) (y + xy')

cos x (1 + sin y) + sin x (cos y) y' = -y sin(xy) - [x sin(xy)]y'

cos x (1 + sin y) + [sin x cos y + x sin(xy)] y' = -y sin(xy)

[sin x cos y + x sin(xy)] y' = -y sin(xy) - cos x (1 + sin y)

Ans. y' = -[y sin(xy) + cos x (1 + sin y)]/[sin x cos y + x sin(xy)]