Question

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

Answer 1

`dy/dx=\frac{ycos(xy)}{1-xcos(xy)}`

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,

`dy/dx=cos(xy) * d/dx(x y)`

`=cos(xy)*(y+x*dy/dx)=ycos(xy)+xcos(xy) dy/dx`

No rearrange this equation as `dy/dx-xcos(xy) dy/dx = ycos(xy)`, factor out the `dy/dx` on the left-hand side and then divide both sides by `1-xcos(xy)` to get

`dy/dx=\frac{ycos(xy)}{1-xcos(xy)}`

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

Answer 2

BTW, the graph of this equation is very beautiful

Explanation:

Here's the graph: