# Question #ef682

Sep 12, 2017

Use implicit differentiation to solve for the derivative to get $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 - 2 x - y \cos \left(x y\right)}{1 + x \cos \left(x y\right)}$.

#### Explanation:

The given equation is $\sin \left(x y\right) + {x}^{2} = x - y$.

We assume this equation defines $y$ as a function of $x$ (you can even write $f \left(x\right)$ in place of $y$ if you want).

Differentiating with respect to $x$ and using this assumption along with the Chain Rule and Product Rule gives:

$\cos \left(x y\right) \cdot \left(1 \cdot y + x \cdot \frac{\mathrm{dy}}{\mathrm{dx}}\right) + 2 x = 1 - \frac{\mathrm{dy}}{\mathrm{dx}} .$

Now multiply this out and rearrange to get the $\frac{\mathrm{dy}}{\mathrm{dx}}$ terms on the left side:

$\left(x \cos \left(x y\right) + 1\right) \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 1 - y \cos \left(x y\right) - 2 x$.

Now divide to get the answer:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 - 2 x - y \cos \left(x y\right)}{1 + x \cos \left(x y\right)}$

BTW, the graph of the given curve is pretty "wild". It is shown below: