Question

Find (d^2y/dx^2) in terms of x and y. 1-x^2y=x-y

Answer 1

@parkjc06 where r u getting confused at?

Answer 2

is it just asking me for the derivative of both sides?

Answer 3

you have to take the derivative in both sides
it is a double derivative

Answer 4

?

Answer 5

\[\frac{ d }{ dx } (1-x^2y) = \frac{ d }{ dx } (x-y)\]
\[0-2*x*y-x^2\frac{ dy }{ dx } = 1 - \frac{ dy }{ dx }\]

Answer 6

do u follow the steps?

Answer 7

after this step u need to derivate again with respect to x

Answer 8

so the second derivative would be 1-2x=0 right?

Answer 9

no
first of all u need to simplify the above expression\[\frac{ dy }{ dx } - x^2\frac{ dy }{ dx } = 1 +2*x*y\]
\[\frac{ dy }{ dx }(1-x^2) = 1 +2*x*y\]

Answer 10

now take the second derivative and apply product rule wherever necessary

Answer 11

well to complete the first derivative it would by dy/dx=[1+2xy]/[1-x^2]

Answer 12

wouldn't that be quotient rule to find second derivative

Answer 13

so you would end up with
y"=(1)(2*dy/dt)-(1+2xy)(-2X)/(1-x^2)?