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

2 Answers
Jun 17, 2015

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

Explanation:

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

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

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

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

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

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

Nov 4, 2017

BTW, the graph of this equation is very beautiful

Explanation:

Here's the graph:

enter image source here