What is the derivative of y=x+sin(xy) in terms of x and y at any point on the curve
So I got to y'\1+cos(xy)*....
I dont know what to do next
2 answers
Sorry that is suppose to say y'=
y=x+sin(xy)
You will need the product rule for the sine term
dy/dx = 1 + cos(xy)(xdy/dx + y)
the second dy/dx is imbedded in a product so we have to expand
I will use y' for dy/dx
y' = 1 + xcos(xy)*y' + ycos(xy)
y' - xcos(xy) y' = 1 + ycos(xy)
y' (1 - xcos(xy) ) = 1 + ycos(xy)
y' = (1 + ycos(xy))/(1 - xcos(xy) )
check: https://www.wolframalpha.com/input/?i=derivative+of+y%3Dx%2Bsin(xy)
You will need the product rule for the sine term
dy/dx = 1 + cos(xy)(xdy/dx + y)
the second dy/dx is imbedded in a product so we have to expand
I will use y' for dy/dx
y' = 1 + xcos(xy)*y' + ycos(xy)
y' - xcos(xy) y' = 1 + ycos(xy)
y' (1 - xcos(xy) ) = 1 + ycos(xy)
y' = (1 + ycos(xy))/(1 - xcos(xy) )
check: https://www.wolframalpha.com/input/?i=derivative+of+y%3Dx%2Bsin(xy)