How do you differentiate #xy=cot(xy)#?

1 Answer
Jul 23, 2015

# dy/dx = - y/x #

Explanation:

I am assuming that you want to find # dy/dx #.

We shall use both implicit differentiation and chain rule.
# xy = cot(xy) #

Differentiate both sides with respect to #x# to get:
# d/dx (xy) = d/dx (cot(xy)) #

Apply chain rule
# d/dx (xy) = -csc^2(xy) d/dx (xy) #
# => (1 + csc^2(xy)) d/dx (xy) = 0 #
# => d/dx (xy) = 0 # (Since #(1 + csc^2(xy))# is always positive)

Apply chain rule again
# y + x dy/dx = 0 #
# dy/dx = - y/x #