Question

how do i find the tenth derivatve of y=1/(2x-1)?

Answer 1

lol your kidding right?

Answer 2

maybe if you take successive derivatives you will see a pattern
why do you need the tenth derivative?

Answer 3

\[y=(2x-1)^{-1}\] usually it is not helpful to use exponential notation but it probably is here
\[y'=-2(x-1)^{-2}\]
\[y''=4(2x-1)^{-3}\]
\[y'''=-12(2x-1)^{-4}\] hmmm looks promising

Answer 4

more promising if i had not made a typo
\[y'=-2(2x-1)^{-1}\]

Answer 5

need something like \((-1)^n\) to make in alternate and also \(n!\) or maybe \(2\times n!\)