Asked by Tim
If x^3 - y^3 =3 , use only implicit diff. to find d^2y/dx^2 . Express your answer in terms of x and y only and write is as a proper fraction that is completely simplified as much as possible.
d^2y/dx^2= -6x/y^5 (this is the answer I should get but i don't know how to show steps for this question)
d^2y/dx^2= -6x/y^5 (this is the answer I should get but i don't know how to show steps for this question)
Answers
Answered by
Steve
-6x/y^5 is not the answer
3x^2 - 3y^2 y' = 0
now use the quotient rule:
y" = (2xy^2 - 2x^2y y')/y^4
= 2x(y-xy')/y^3
= 2x(y-x(x^2/y^2))/y^3
= 2x(y^3-x^3)/y^5
or, use implicit differentiation twice:
x^3-y^3 = 3
3x^2 - 3y^2 y' = 0
6x - 6y(y')^2 - 3y^2y" = 0
y^2 y" = 2x - 2y(y')^2
= 2(x -y(x^2/y^2)^2)
= 2(xy^4-x^4y)/y^4
= 2x(y^3-x^3)/y^3
y" = 2x(y^3-x^3)/y^5
3x^2 - 3y^2 y' = 0
now use the quotient rule:
y" = (2xy^2 - 2x^2y y')/y^4
= 2x(y-xy')/y^3
= 2x(y-x(x^2/y^2))/y^3
= 2x(y^3-x^3)/y^5
or, use implicit differentiation twice:
x^3-y^3 = 3
3x^2 - 3y^2 y' = 0
6x - 6y(y')^2 - 3y^2y" = 0
y^2 y" = 2x - 2y(y')^2
= 2(x -y(x^2/y^2)^2)
= 2(xy^4-x^4y)/y^4
= 2x(y^3-x^3)/y^3
y" = 2x(y^3-x^3)/y^5
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