Question #36537

1 Answer
Oct 30, 2016

#(d^2y)/dx^2 = (3x^2(y^4-x^4))/y^7 = (-3x^2)/y^7#

Explanation:

#x^4-y^4=1#

Differentiate both sides of the equation with respect to #x#

#4x^3 - 4y^3 dy/dx = 0#

So #dy/dx = x^3/y^3#

Now differentiate again w.r.t. #x#.

#(d^2y)/dx^2 = (3x^2y^3-x^3 3y^2 dy/dx)/y^6#

Get rid of the common factor #y^2#

# = (3x^2y-3x^3dy/dx)/y^4#

Replace #dy/dx = x^3/y^3#

# = (3x^2y-3x^3 x^3/y^3)/y^4#

Clear the fraction in the numerator by multiplying by #y^3/y^3#

# = (3x^2y^4-3x^6)/y^7#

Factor the numerator.

# = (3x^2(y^4-x^4))/y^7#

Recognize the opposite of the initial expression.

#x^4-y^4=1 iff y^4-x^4=-1#