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

2 Answers
Jun 17, 2015

Answer:

#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.

Nov 4, 2017

Answer:

BTW, the graph of this equation is very beautiful

Explanation:

Here's the graph:

enter image source here