Posted by Alan on .
Find the equation of the tangent line to the graph of √(xy)=x2y at (4,1)

Calculus 
Reiny,
(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
y1 = (1/2)(x4)
2y2= x+4
x + 2y = 6
or
y = (1/2)x + 6