Question

find the first and second derivatives of x^2 + xy - y^2 = 4 , express the second derivative in terms of x and y

Answer 1

2x + x(dy/dx) + y - 2y(dy/dx) = 0 ???is this right?

Answer 2

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

Answer 3

wolfram says yes, now how do I find the second derivative

Answer 4

looks like I need to substitute the first derivative for instances of y' in the second derivative, is that it?

Answer 5

\[\frac{ dy }{ dx }=\frac{ y+2x }{ 2y-x }\]
\[\frac{ d^2y }{ dx^2 }=\frac{ \left( y-2x \right)\left( \frac{ dy }{ dx }+2 \right)-\left( y+2x \right)\left( 2 \frac{ dy }{ dx }-1 \right) }{ \left( 2y-x \right)^2}\]
put the value of dy/dx and simplify.