Find dy/dx and d^2y/dx^2 at the point (2,1) on the curve x^2 + 2xy - y^2=7.?

Question

Show a lot of work please.

Answer 1

x^2+2xy-y^2=7

dy/dx
2x+2y+2x(dy/dx)-2(dy/dx)=0
(2x-2)(dy/dx)=-2x-2y

dy/dx= (-2x-2y)/(2x-2)
dy/dx=-3

d^2y/dx^2
[(2x-2)(-2-2(dy/dx))-
(-2x-2y)(2)]/(2x-2)^2

d^2y/dx^2=5

i think thats rite sry bout all the parantheses

Answer 2

I find it easier to do it by taking the derivative of both sides of the equation:
d/dx[ equation above] = 2x - 2y - 2y(dy/dx) = 0
Divide both sides by two and solve for dy/dx:
-y(dy/dx) = y-x
dy/dx = (x-y)/y; now plug in (2,1) and you are done for this part

d2y/dx2 = (d/dx)[(x-y)/y] = (d/dx)(x/y) - (d/dx)(1)
= 1/y - 0; again plug in (2,1) and you are el done (Spanish for done)