Calculus! Help!?

Question

find y' of y*sqrt(x-1) + x*sqrt(y-1) = xy

Please Explain Steps.

Answer 1

Use implicit differentialtion. Take the derivative d/dx of all terms

d/dx { y*√[x-1] } = y'*√[x-1] + (1/2)*(1/√[x-1])*y

d/dx { x*√[y-1] } = √[y-1] + x*(1/2)*(1/√[y-1])*y'

d/dx { xy } = x*y' + y

Put it all together and solve for y'

Answer 2

look at the problem as 3 simple problems and differentiate them using chain rule:
1. d(y*sqrt(x-1))=[y'*(sqrt(x-1)]+[(y/2)*(x-1)^(-0.5)]
2. d(x*sqrt(y-1))=[(sqrt(y-1)]+[((x*y')/2)*(y-1)^(-0.5)]
3. d(xy)= (x*y')+y
Now just put the differentiated parts back in the original equation and solve for y'

Answer 3

Differentiate with respect to x,
y'√(x-1) + y/[2√(x-1)] + √(y-1) + xy'/[2√(y-1)] = y + xy'
Solving for y' will give you the answer.

Answer 4

Find the deriviative dy/dx of:

y√(x - 1) + x√(y - 1) = xy

Differentiate implicitly.

y/[2√(x - 1)] + (dy/dx)√(x - 1) + √(y - 1) + x(dy/dx)/[2√(y - 1)]
= y + dy/dx

(dy/dx){√(x - 1) + x/[2√(y - 1)] - 1} = y - y/[2√(x - 1)] - √(y - 1)

dy/dx = {y - y/[2√(x - 1)] - √(y - 1)} / {√(x - 1) + x/[2√(y - 1)] - 1}

Answer 5

Hi,

you take it as d/dx(uv) formula.

http://www.pen.k12.va.us/Div/Winchester/jhhs/math/lessons/calculus/deriv2.html