# How do you differentiate y=sin(xy)?

Jun 17, 2015

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

#### Explanation:

Assuming you are differentiating with respect to $x$ and assuming that this equation implicitly defines $y$ as a function of $x$, you get, by using the Chain Rule and Product Rule,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \cos \left(x y\right) \cdot \frac{d}{\mathrm{dx}} \left(x y\right)$

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

No rearrange this equation as $\frac{\mathrm{dy}}{\mathrm{dx}} - x \cos \left(x y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = y \cos \left(x y\right)$, factor out the $\frac{\mathrm{dy}}{\mathrm{dx}}$ on the left-hand side and then divide both sides by $1 - x \cos \left(x y\right)$ to get

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

Since the original equation cannot be solved explicitly for $y$ as a function of $x$, this is the best you can do.

Nov 4, 2017

BTW, the graph of this equation is very beautiful

#### Explanation:

Here's the graph: