Verified answer

To differentiate sqrt(2xy) with respect to x you can write

sqrt(2xy) = (2xy)^(1/2)

and then use the chain rule

d/dx (2xy)^(1/2) = (1/2)*(2xy)^(-1/2) * d/x (2xy)

and then the chain rule

d/dx (2xy) = 2y + 2x dy/dx

OK, so that takes care of the part you said you're concerned about, and just for completeness the whole thing is

0 = d/dx sqrt(2xy) - 1 - 3y^2

= (1/2) (2xy)^(-1/2) (2y + 2x dy/dx) - 0 + 6y dy/dx

and solving for dy/dx you get

dy/dx = -y (2xy)^(-1/2) / (6y + x (2xy)^(-1/2))

First off, rewrite the square root as (2xy)^1/2. It'll make things easier.

Differentiate it like you normally would (using a combination of Chain Rule and Product Rule), but instead of the derivative of y just being 1, it's going to be y'. So...

[(1/2)(2xy)^(-1/2)]*(2y + 2x*y') = 6y*y'

you mean root ( 2xy ) ?

easier to think of it with the constant outside.

root(2) * (xy )^(1/2)

now take the derivative

root(2) * [ 1/2 * (xy)^(-1/2) * (y + x * dy/dx ) ]

the first is just the constant, the second part just inside the brackets is the first step of the derivative of the product, the second part inside the brackets is from the chain rule of the product (xy)

The gist of the communicate accessible (sorry, there are too many pages to supply all the links) is that wands help to concentration the magic resident interior the witch or wizard, yet wandless magic remains obtainable, extremely for terribly proficient people like Snape and Dumbledore. some examples ... Apparition Assuming one's Animagus style (or Metamorphpagus) Accio (the Summoning allure) Elves can do magic with out making use of wands Lumos

(√2xy) = (2xy)^(1/2) = 2^(1/2) * x^(1/2) * y^(1/2)

(√2xy) - 1 = 3y^2

2^(1/2) * (1/2)x^(-1/2) * y^(1/2) + 2^(1/2) * x^(1/2) * (1/2)y^(-1/2) * y' = 6y

2^(1/2) * x^(1/2) * (1/2)y^(-1/2) * y' = 6y - 2^(1/2) * (1/2)x^(-1/2) * y^(1/2)

y' = [6y - 2^(1/2) * (1/2)x^(-1/2) * y^(1/2)] / [2^(1/2) * x^(1/2) * (1/2)y^(-1/2)]

√2y + √2x*dy/dx = 6y*dy/dx

6y*dy/dx - √2x*dy/dx = √2y

dy/dx = [ √2y ] / [ 6y - √2y ]