Question #26eaf

2 Answers
Jan 30, 2018

#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!

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!