Question

please explain this: (problem's in the comments section) Thanks.

Answer 1

Find \[\frac{ d^2y }{ dx^2 }\] of y=x/1-x

Answer 2

\[y=\frac x{1-x}?\]

Answer 3

Yes. :)

Answer 4

\[y=\frac{x-1}{1-x}+\frac1{1-x}=-1+\frac1{1-x}\]

Answer 5

\[\frac{dy}{dx}=\frac d{dx}\frac1{1-x}=\frac1{(1-x)^2}\]

Answer 6

Yes, that's the derivative, but then, how to get the d^2? I don't know that one.

Answer 7

It's like finding the derivative's derivative

Answer 8

What do I do if the d has an exponent?

Answer 9

\[\frac{d^2y}{(dx)^2}=\frac d{dx}\frac{dy}{dx}=\frac d{dx}\frac1{(1-x)^2}\]

Answer 10

Just a notation

Answer 11

Oh, oh I got it. Thanks!

Answer 12

np :)