Question

find second derivative if y =(1-x)/(2-x)

Answer 1

u know Quotient Rule..?

Answer 2

i got 1st derivative as -1/(x^2-4x+4)...if thats even right...

Answer 3

Yah I think that's correct.
You don't need to expand out the bottom though, just leave it as a square.

Answer 4

\[\frac{ dy }{ dx }=\frac{ -1 }{ x ^{2}-4x+4 }\]

Answer 5

or,\[\frac{ d ^{2}y }{ dx ^{2} }=\frac{ (-1)(-1) (2x-4)}{(x ^{2}-4x+4)^{2} }\]

Answer 6

did u get my solution

Answer 7

answer is (-2)/(2-x)^3
.......wish i know how to get that..

Answer 8

\[\large y=\frac{(1-x)}{(2-x)}\]\[\large y'=\frac{(1-x)'(2-x)-(1-x)(2-x)'}{(2-x)^2}\]
\[\large y'=\frac{(-1)(2-x)-(1-x)(-1)}{(2-x)^2} \quad =\quad \frac{-1}{(2-x)^2}\]
\[\large y'=-(2-x)^{-2}\]Don't apply quotient rule for this next one, simply do power rule.\[\large y''=-(-2)(2-x)^{-3}(-1) \quad = \quad \frac{-2}{(2-x)^3}\]