Question #26eaf

2 Answers
Jan 30, 2018

dy/dx = y/(x(2y^2-1))dydx=yx(2y21)

Explanation:

y^2 = lnxyy2=lnxy

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

y^2 = lnx + lnyy2=lnx+lny

Then use implicit differentiation to differentiated with respect to xx. 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!

Jan 30, 2018

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!