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Show that d^2y/dx^2=-2x/y^5, if x^3 + y^3=1

User Lida
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1 Answer

7 votes

Answer:

y³ + x³ = 1

First, differentiate the first time, term by term:


{3y^(2).(dy)/(dx) + 3x^(2)} = 0 \\\\{3y^(2).(dy)/(dx) = -3x^(2)} \\\\(dy)/(dx) = (-3x^(2))/(3y^(2)) \\\\(dy)/(dx) = (-x^(2))/(y^(2))

↑ we'll substitute this later (4th step onwards)

Differentiate the second time:


3y^(2).(dy)/(dx) + 3x^(2) = 0 \\\\3y^(2).(d^(2) y)/(dx^(2)) + 6y((dy)/(dx))^(2) + 6x = 0 \\\\3y^(2).(d^(2) y)/(dx^(2)) + 6y((dy)/(dx))^(2) = - 6x \\\\3y^(2).(d^(2) y)/(dx^(2)) + 6y((-x^(2) )/(y^(2) ))^(2) = - 6x \\\\3y^(2).(d^(2) y)/(dx^(2)) + 6y((x^(4) )/(y^(4) )) = - 6x \\\\3y^(2).(d^(2) y)/(dx^(2)) + (6x^(4) )/(y^(3) ) = - 6x \\\\3y^(2).(d^(2) y)/(dx^(2)) = - 6x - (6x^(4) )/(y^(3) ) \\\\


3y^(2).(d^(2) y)/(dx^(2)) = - (- 6xy^(3) - 6x^(4) )/(y^(3)) \\\\(d^(2) y)/(dx^(2)) = - (- 6xy^(3) - 6x^(4) )/(3y^(2). y^(3)) \\\\(d^(2) y)/(dx^(2)) = - (- 2xy^(3) - 2x^(4) )/(y^(5)) \\\\(d^(2) y)/(dx^(2)) = - (-2x (y^(3) + x^(3)))/(y^(5)) \\\\(d^(2) y)/(dx^(2)) = - (-2x (1))/(y^(5)) \\\\(d^(2) y)/(dx^(2)) = - (-2x)/(y^(5))

User Aadu
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