Question

Find the equation of the tangent line to the graph of √(xy)=x-2y at (4,1)

Answer 1

(1/2)(xy)^(-1/2) (x dy/dx + y) = 1 - 2dy/dx
at the given point

(1/2)(4)^(-1/2) (4dy/dx + 1) = 1 - 2dy/dx
(1/4)(4dy/dx + 1 = 1 - 2dy/dx
times 4
4dy/dx + 4 = 1 - 2dy/dx
6dy/dx = -3
dy/dx = -1/2

y-1 = -(1/2)(x-4)
2y-2= -x+4
x + 2y = 6
or
y = (-1/2)x + 6