Question

What is the second derivative of x^2-xy+y^2=1?

Answer 1

The first derivative is: y'=(y-2x)/(2y-x)
I don't know how to find the 2nd derivative, but I do know the 1st derivative is needed to plug in for y'.
Please help?

Answer 2

The second derivative is just the derivative of the first derivative.

Answer 3

im also studying for derivatives, but how did you even get to the first derivative?!?! :O

Answer 4

Here's what I got\[x^2 - xy + y^2 = 1\] Differentiating a first time,\[ \frac{ d }{ dx }(x^2 -xy + y^2 = 1)\]\[\rightarrow 2x - xy' - y + 2yy' = 0\]\[y'(2y-x) = y - 2x\]\[y' = \frac{ y - 2x }{ 2y - x }\]

Answer 5

Do you understand how implicit differentiation works?

Answer 6

Well, it's pretty much using the chain rule on the dependent variable. In this case, it's y.
\[\frac{ d }{ dx }(f(g(x)) = f'(g(x))g'(x)\]\[\frac{ d} {dx}(y(x)) = y'(x) x' = (y')(1)\]So, the derivative of \[y^2\]Would be \[2y y'\]

Answer 7

Hi, I do know implicit differentiation and chain rules, but I am having trouble calculating it numerically! Can someone PLEASE show me step by step how to do it??

The answer is supposed to be:
-6/((2y-x)^3) but I don't know how to get it!

Answer 8

\(x^2-xy+y^2=1\)

\(2x - (xy' + y) + 2yy' = 0\)

\(2x - xy' - y + 2yy' = 0\)

\(2yy' - xy' = y - 2x\)

\(y'(2y - x) = y - 2x\)

\(y' = \dfrac{y - 2x}{2y - x} \)

\(y'' = \dfrac{(2y - x)(y' - 2) - (y - 2x)(2y' - 1)}{(2y - x)^2 } \)

Now you can substitute y' with what we found y' to be above and simplify.

Answer 9

Hi! This is what I have done, but when I substitute y' in, it doesn't work! Can you try putting y' in it to show me how it's done?

The answer is supposed to be: -6/((2y-x)^3) !

Answer 10

\(y'' = \dfrac{(2y - x)(\dfrac{y - 2x}{2y - x} - 2) - (y - 2x)(2 \cdot\dfrac{y - 2x}{2y - x} - 1)}{(2y - x)^2 }\)

\(y'' = \dfrac{y - 2x - 4y + 2x - (2y - 4x)(\dfrac{y - 2x}{2y - x} - 1)}{(2y - x)^2 }\)

\(y'' = \dfrac{ - 3y - \left(\dfrac{2y^2 - 4xy - 4xy + 8x^2}{2y - x} - 2y + 4x \right)}{(2y - x)^2 }\)

\(y'' = \dfrac{-3x - \dfrac{2y^2 - 8xy + 8x^2}{2y - x} + 2y - 4x }{(2y - x)^2 }\)

\(y'' = \dfrac{ \dfrac{ -3x(2y - x)}{2y - x} - \dfrac{2y^2 - 8xy + 8x^2}{2y - x} + \dfrac{(2y - 4x)(2y - x)}{2y - x} }{(2y - x)^2 }\)

\(y'' = \dfrac{ -3x(2y - x) - (2y^2 - 8xy + 8x^2) + (2y - 4x)(2y - x)}{(2y - x)^3 }\)

\(y'' = \dfrac{ -3x(2y - x) - (2y^2 - 8xy + 8x^2) + 4y^2 - 2xy -8xy + 4x^2}{(2y - x)^3 }\)

\(y'' = \dfrac{ -6xy + 3x^2 - 2y^2 + 8xy - 8x^2 + 4y^2 -10xy + 4x^2}{(2y - x)^3 }\)

\(y'' = \dfrac{ -x^2 -2xy + 6y^2 }{(2y - x)^3 }\)

Answer 11

mathstudent55
you messed up on your second step
you cannot distribute the 2 to the y-2x because then it will eventually changes your -1 at the end
\[(y-2x)(2*\frac{ y-2x }{ 2y-x }-1)\]
it would have to end up being 1/2

Answer 12

\(y'' = \dfrac{(2y - x)(\dfrac{y - 2x}{2y - x} - 2) - (y - 2x)(2 \cdot\dfrac{y - 2x}{2y - x} - 1)}{(2y - x)^2 }\)

Answer 13

\(y'' = \dfrac{y - 2x - 4y + 2x - (y - 2x)(\dfrac{2y - 4x}{2y - x} - 1)}{(2y - x)^2 }\)

Answer 14

\(y'' = \dfrac{-3y - \dfrac{2y^2 - 4xy - 4xy +8x^2}{2y - x} + y - 2x}{(2y - x)^2 }\)

Answer 15

\(y'' = \dfrac{ ~~~\dfrac{-3y(2y - x) - 2y^2 + 8xy - 8x^2 + (y - 2x)(2y - x)}{2y - x} ~~~~}{(2y - x)^2} \)

Answer 16

\(y'' = \dfrac{ -6y^2 + 3xy - 2y^2 + 8xy - 8x^2 + 2y^2 -5xy + 2x^2}{(2y - x)^3}\)

Answer 17

\(y'' = \dfrac{ -6y^2 + 6xy -6x^2}{(2y - x)^3}\)

Answer 18

\(y'' = \dfrac{ -6(y^2 - xy +x^2)}{(2y - x)^3}\)

Answer 19

@hasalotoftrouble
You are correct. I treated it as if there were a close parentheses after the 2.

Answer 20

i have the same problem except i can't come out with \[\frac{ -6 }{ (2y-x)^3 }\]

Answer 21

I redid it now by distributing the 2 by the numerator of the fraction after the 2.
It's late, and I am not sure I did all the algebra correctly this time either.

Answer 22

Look at my new solution. I do have -6 in the numerator multiplying a trinomial.

Answer 23

i keep coming out with the answer you have but can't do anything with it

Answer 24

Then maybe my algebra is correct despite the time.

Answer 25

i see that but how do you get rid of the trinomial

Answer 26

Thanks for pointing out my mistake.
The trinomial is not factorable, so I can't do anything with it.

Answer 27

so either we are both incorrect or the book i have is wrong
(the book is never wrong)

Answer 28

especially not a AP book

Answer 29

i found out how to make part of the numerator go away

Answer 30

the original equation is in our second derivatives numerator

Answer 31

and it is equal to 1 making the numerator -6(1)

Answer 32

then you get the answer\[\frac{ -6 }{ (2y-x)^3 }\]

Answer 33

credit to satellite73

Answer 34

@hasalotoftrouble
Great! Thanks for that. Now I see it.