Question

What is the implicit derivative of `1= xye^y-xcos(xy)`?

Answer 1

`(dy)/(dx) = (cos(xy) - xysin(xy) - ye^y)/(xe^y+xye^y+x^2sin(xy))`

Explanation:

`d/(dx)(1) = d/(dx)(xye^y) - d/(dx)(xcos(xy))`

To do these derivatives, remember `d/(dx)(y) = (dy)/(dx)` and hammer your product and chain rules.

`0 = ye^y + x(dy)/(dx)e^y + xy(dy)/(dx)e^y - cos(xy) + x(y+x(dy)/(dx))sin(xy)`

`(dy)/(dx)(xe^y+xye^y+x^2sin(xy)) = cos(xy) - xysin(xy) - ye^y`

`(dy)/(dx) = (cos(xy) - xysin(xy) - ye^y)/(xe^y+xye^y+x^2sin(xy))`