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

1 Answer
Jul 23, 2015

dy/dx = - y/x dydx=yx

Explanation:

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

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

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

Apply chain rule
d/dx (xy) = -csc^2(xy) d/dx (xy) ddx(xy)=csc2(xy)ddx(xy)
=> (1 + csc^2(xy)) d/dx (xy) = 0 (1+csc2(xy))ddx(xy)=0
=> d/dx (xy) = 0 ddx(xy)=0 (Since (1 + csc^2(xy))(1+csc2(xy)) is always positive)

Apply chain rule again
y + x dy/dx = 0 y+xdydx=0
dy/dx = - y/x dydx=yx