Question

Question #26eaf

Answer 1

`dy/dx = y/(x(2y^2-1))`

Explanation:

`y^2 = lnxy`

Using the logarithm law that the logarithm of a product is equal to the sum of the two logarithms:

`y^2 = lnx + lny`

Then use implicit differentiation to differentiated with respect to `x`. Remember that using the chain rule `d/dx(f(y)) = f'(y)*dy/dx`.

Hence, `2y*dy/dx = 1/x + 1/y*dy/dx`

Factorise out `dy/dx` next!

`dy/dx(2y - 1/y) = 1/x`

A bit of algebraic juggling shows that `2y - 1/y = (2y^2-1)/y`.

Hence, `dy/dx = 1/x -: (2y^2-1)/y`

`dy/dx = y/(x(2y^2-1))`

And we're done!

Answer 2

`dy/dx=y/{x(2y^2-1)}`

Explanation:

`y^2=ln(xy) rArr xy=e^(y^2)rArr x=e^(y^2)/y............(square)`.

Diff.ing w.r.t. `y`, using the Quotient Rule, we have,

`dx/dy={(yd/dye^(y^2)-e^(y^2)d/dyy)}/y^2`,

`={y*e^(y^2)*d/dyy^2-e^(y^2)*1}/y^2`,

`={y*e^(y^2)*2y-e^(y^2)*1}/y^2`,

`={e^(y^2)*(2y^2-1)}/y^2`,

`={(xy)(2y^2-1)}/y^2`,

`rArr dy/dx=1/{dx/dy}=y/{x(2y^2-1)}`, is the desired derivative!