Question

let y=1/7x then

(1/n!)(d^ny/dx^n)=

Answer 1

\[y = \frac{ 1 }{ 7x } = \frac{ 1 }{ 7 } * \frac{ 1 }{ x } = \frac{ 1 }{ 7 } * x^{-1}\] \[\frac{ dy }{ dx } = \frac{ 1 }{ 7 } * (-1)(x)^{-2}\]\[\frac{ d^{2}y }{ dx^{2} } = \frac{ 1 }{ 7 } * (-1)(-2)(x)^{-3}\]\[\frac{ d^3y }{ dx^3 } = \frac{ 1 }{ 7 } * (-1)(-2)(-3)(x)^{-4}\]...
\[\frac{ d^ny }{ dx^n } = \frac{ 1 }{ 7 } * (-1)(-2)...(-n)(x)^{-(n+1)}\]
\[\frac{ d^n y }{ dx^n } = (-1)^n \frac{ n! }{ 7x^{n+1} }\]
\[\frac{ 1 }{ n! } * \frac{ d^n y }{ dx^n } = \frac{ 1 }{ 7x^{n+1} }\]
?

Answer 2

the last line should be \[(-1)^n \frac{ 1 }{ 7x^{n+1} }\]

Answer 3

or just
\[\frac{ (-1)^n }{ 7x^{n+1} }\]
?

Answer 4

@binarymimic why do you have two answers?

Answer 5

where?

Answer 6

so the answer is \[\frac{ (-1)^n }{ 7x ^{n+1} }\]

Answer 7

that is not right though

Answer 8

i wonder why? do you have the correct answer? seems like it works for any n

Answer 9

can anyone do this?

Answer 10

@jim_thompson5910

Answer 11

@primeralph