Question

find the second derivative of y^2=x^2+2x

Answer 1

just derive it like you normally would, implicits aint a new thing :)

Answer 2

\[1/y^3\] is this the answer

Answer 3

i doubt it

Answer 4

what steps did you take to derive it?

Answer 5

differentiating both sides wrt x
then double derivative

Answer 6

ok, what is the first derivative wrtx of:

y^2=x^2+2x

Answer 7

(x+1)/y

Answer 8

2y y' = 2x + 2

y' = (x+1)/y ; yep, i agree

and the derivative of that is?

Answer 9

\[(y-(x+1)dy/dx)/y ^{2}\]

Answer 10

y'' = [y(x+1)' - y'(x+1)]/y^2

y'' = (y - xy'+y')/y^2

y'' = (y^2/y - x(x+1)/y +(x+1)/y)/y^2

y'' = (y^2 - x(x+1) +(x+1))/y^3

y'' = (y^2 - x^2-x+x+1)/y^3

y'' = (y^2 - x^2 +1)/y^3

is what I get

Answer 11

forgot to drag the negative thru in the second step tho

Answer 12

y'' = [y(x+1)' - y'(x+1)]/y^2

y'' = (y - xy'-y')/y^2

y'' = (y^2/y - x(x+1)/y -(x+1)/y)/y^2

y'' = (y^2 - x(x+1) -(x+1))/y^3

y'' = (y^2 - x^2-x-x-1)/y^3

y'' = (y^2 - x^2-2x-1)/y^3

looks a bit better maybe

Answer 13

can't we put value of y^2 from the question? if so the ans is 1/y^3