what is dy/dx of x^2 + sin(xy) +y^2 = 0?
I would guess the answer is
2x + cos (xy)* y = 0
Is this correct?
MH
First, please stop saying "dy/dx of" something. You are asking for dy/dx itself, where y is (implicitely) defined by that equation.
However yes, the derivative of sin(u) with respect to x is cos(u)(du/dx). And the derivative of u= xy, with respect to x, is (dx/dx)y+ x(dy/dx)= y+ xdy/dx so that the derivative of sin(xy), with respect to x, is cos(xy)(y+ x dy/dx).
This is because $\displaystyle \displaystyle \begin{align*} \frac{d}{dx} \end{align*}$ is the OPERATOR, while $\displaystyle \displaystyle \begin{align*} \frac{dy}{dx} = \frac{d}{dx} \left( y \right) \end{align*}$, the RESULT after the operator has operated on the function y.