Implicitly differentiate $e^y \cos(x) = 1 + \sin(xy)$

Question

I can differentiate one side of the equation, but I dont know how to deal with sin(xy)

Answer 1

Hint: use the change rule (sin(xy))'(x) = cos(xy)*(y + xy') and you can continue:

Answer 2

Are you differentiating "with respect to x" or "with respect to y"? In other words, are you differentiating the function defined by $f_1(x) = e^{g(x)} \cos(x) = 1 + \sin(xg(x))$ or $f_2(y) = e^y \cos(h(y)) = 1 + \sin(h(y)y)$. If its the first one, then we have $$ {f_1}'(x) = e^{g(x)}g'(x) \cdot \cos(x) - e^{g(x)} \sin(x) = \cos(xg(x)) \cdot \left(g(x) + xg'(x)\right). $$ If its the second one, then we have $$ {f_2}'(y) = e^y \cos(h(y)) - e^y \sin(h(y)) \cdot h'(y) = \cos(h(y)y) \cdot \left(h'(y)y + h(y)\right). $$