use the implicit differentiation to find the derivative of xy=cot(xy)?

Answer 1

Key things to know for implicit differentiation: chain rule, and noting that the derivative of y is written as y'.

xy = cot(xy)

First, differentiate both sides. On the left hand side, we use the product rule and then the chain rule, and on the right hand side, we use the chain rule and then the product rule.

(1)y + xy' = -csc^2(xy) [y + xy']
y + xy' = -csc^2(xy) [y + xy']

Incorrect step: do NOT divide both sides by y + xy', because it eliminates what we want to solve for (which is y'). Our goal here is to isolate y'. Instead, we expand the right hand side.

y + xy' = -y csc^2(xy) - xy' csc^2(xy)

Move all terms with y' to the left hand side, and move all terms without y' to the right hand side.

xy' + xy'(csc^2(xy)) = -y csc^2(xy) - y

Factor out a y',

y' (x + x csc^2(xy)) = -y csc^2(xy) - y
y' = [-y csc^2(xy) - y]/[x + x csc^2(xy)]
y' = [ -y (csc^2(xy) + 1) ] / [x (1 + csc^2(xy))]
y' = [-y/x] (csc^2(xy) + 1)/(1 + csc^2(xy))

Note that, aside from the order, we have the same term on the top and bottom, so they cancel out to 1.
y' = [-y/x]

Answer 2

Using the chain rule, xy differentiates to y + xy'.
Again using the chain rule, cot(xy) differentiates to -cosec(xy)^2 * (y + xy').

So we have y+xy' = -cosec(xy)^2 * (y+xy')
So (y+xy')(1+cosec(xy)^2) = 0
y + xy' = 0 (second bracket is > 0)
y' = -y/x.

Answer 3

4xy - 3x^2 = 0 4xy = 3x^2 y = (3x^2) / (4x) y = (3x) / 4, or x = 0 If x = 0, y must be some thing. that is a vertical line, so the slope (spinoff) is infinity. If y = (3x)/4 = (3/4)x, the spinoff is 3/4 or 0.seventy 5.