# I have the following expression which I need to implicitly differentiate: xy^2+x^2+y+sin(x^2y)=0

I have the following expression which I need to implicitly differentiate:
$x{y}^{2}+{x}^{2}+y+\mathrm{sin}\left({x}^{2}y\right)=0$
I'm a little confused as I'm not entirely sure what to do with the trig function. Here is my work so far:
$\frac{dy}{dx}\left[x{y}^{2}+{x}^{2}+y+\mathrm{sin}\left({x}^{2}y\right)\right]=\frac{dy}{dx}0$
$\frac{d{y}^{2}}{dx}+2x+\frac{dy}{dx}+\mathrm{cos}\left({x}^{2}y\right)\left(2x\frac{dy}{dx}\right)=0$
How should I proceed?
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Simeon Hester
$\frac{d}{dx}\left[x{y}^{2}+{x}^{2}+y+\mathrm{sin}\left({x}^{2}y\right)\right]=\frac{d}{dx}\left(0\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{y}^{2}+2xy\frac{dy}{dx}+2x+\frac{dy}{dx}+\mathrm{cos}\left({x}^{2}y\right)\left(2xy+{x}^{2}\frac{dy}{dx}\right)=0$
We use the product rule and chain rule here, and also the operator for differentiation is
$\frac{d}{dx}$
###### Did you like this example?
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. $y=y\left(x\right)$).
$\begin{array}{rl}\frac{d}{dx}\left[x{y}^{2}+{x}^{2}+y+sin\left({x}^{2}y\right)\right]& =\frac{d}{dx}0\\ {y}^{2}+x\left(2y\right)\frac{dy}{dx}+2x+\frac{dy}{dx}+\mathrm{cos}\left({x}^{2}y\right)\left[\frac{d}{dx}\left({x}^{2}y\right)\right]& =0\\ {y}^{2}+x\left(2y\right)\frac{dy}{dx}+2x+\frac{dy}{dx}+\mathrm{cos}\left({x}^{2}y\right)\left(2xy+{x}^{2}\frac{dy}{dx}\right)& =0\\ \frac{dy}{dx}\left(2xy+1+{x}^{2}\mathrm{cos}\left({x}^{2}y\right)\right)& =-\left({y}^{2}+2x\right)\\ \frac{dy}{dx}& =-\frac{\left({y}^{2}+2x\right)}{\left(2xy+1+{x}^{2}\mathrm{cos}\left({x}^{2}y\right)\right)}\end{array}$