Implicit differentiation with trig function

Question

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?

Answer 1

$$ \frac{d}{dx} [xy^2 + x^2 + y + \sin(x^2y)] = \frac{d}{dx}(0) \\ \implies y^2+2xy\frac{dy}{dx}+2x + \frac{dy}{dx} + \cos(x^2y)(2xy+x^2\frac{dy}{dx})=0$$

We use the product rule and chain rule here, and also the operator for differentiation is $$\frac{d}{dx}$$

Answer 2

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*}

Answer 3

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.