Implicit differentiation with trig function I have the following expression which I need to implicitly differentiate:
$$ xy^2 + x^2 + y + \sin(x^2y) = 0 $$
I'm a little confused as I'm not entirely sure what to do with the trig function.  Here is my work so far:
$$ \frac{dy}{dx} [xy^2 + x^2 + y + \sin(x^2y)] = \frac{dy}{dx}0 $$
$$ \frac{dy^2}{dx} + 2x + \frac{dy}{dx} + \cos(x^2y)(2x\frac{dy}{dx}) = 0 $$
How should I proceed?
 A: $$ \frac{d}{dx} [xy^2 + x^2 + y + \sin(x^2y)] = \frac{d}{dx}(0) \\
\implies y^2+2xy\frac{dy}{dx}+2x + \frac{dy}{dx} + \cos(x^2y)(2xy+x^2\frac{dy}{dx})=0$$
We use the product rule and chain rule here, and also the operator for differentiation is $$\frac{d}{dx}$$
A: First, you should take the derivative of both sides of the equation (apply $\frac{d}{dx}$ instead of $\frac{dy}{dx}$). Then make use of the chain and product rules. It helps to think of $y$ as a function of $x$ (i.e. $y = y(x)$).
\begin{align*}
\frac{d}{dx}[xy^2+x^2+y+sin(x^2y)]&=\frac{d}{dx}0\\
y^2+x(2y)\frac{dy}{dx} + 2x+\frac{dy}{dx}+\cos(x^2y)\left[\frac{d}{dx}(x^2y)\right] &=0\\
y^2+x(2y)\frac{dy}{dx} + 2x+\frac{dy}{dx}+\cos(x^2y)(2xy+x^2\frac{dy}{dx}) &=0\\
\frac{dy}{dx}\left(2xy +1 +x^2\cos(x^2y)\right) &=- (y^2+ 2x) \\
\frac{dy}{dx} &=-\frac{ (y^2+ 2x)}{\left(2xy +1 +x^2\cos(x^2y)\right)} \\
\end{align*}
A: In his answer, CarlOlimb says "It helps to think of y as a function of x" When I was teaching implicit differentiation, I'd suggest to my students to replace y by f(x). Those who did this made less errors than the ones who didn't. Of course, as one becomes more proficient, this replacement becomes less useful. 
