Implicitly differintiate?

Question

implicit differintiate:
y times the square root of x minus x times the square root of y equals 16

Answer 1

y√x - x√y = 16

Differentiating with respect to x:

d(y√x - x√y)/dx = 0
d(y√x)/dx - d(x√y)/dx = 0
dy/dx √x + yd(√x)/dx - (dx/dx √y + x d(√y)/dx) = 0
dy/dx √x + y/(2√x) - √y - x/(2√y) dy/dx = 0

Solving for dy/dx:

dy/dx √x + y/(2√x) - √y - x/(2√y) dy/dx = 0
dy/dx √x - x/(2√y) dy/dx = √y - y/(2√x)
dy/dx (√x - x/(2√y)) = √y - y/(2√x)
dy/dx = (√y - y/(2√x))/(√x - x/(2√y))

And we are done.

Answer 2

y√x - x√y = 16

take derivaties
d/dx (y√x) - d/dx (x√y) = d/dx (16)

product rule:
[√x d/dx(y) + y d/dx(√x)] - [√y d/dx(x) + x d/dx(√y)] = 0

simplify
[√x d/dx(y) + y * 1/(2√x)] - [√y + x * 1/(2√y) dy/dx] = 0

simplify
[√x dy/dx + y/(2√x)] - [√y + x/(2√y) dy/dx] = 0

distribute
√x dy/dx + y/(2√x) - √y - x/(2√y) dy/dx = 0

subtract y/(2√x) and add √y for both sides
√x dy/dx - x/(2√y) dy/dx = √y - y/(2√x)

take out dy/dx
dy/dx [√x - x/(2√y)] = √y - y/(2√x)

now divide
dy/dx = [√y - y/(2√x)] / [√x - x/(2√y)] <== answer

Rec