Question

Find the derivative of y with respect to the appropriate varible.
y=cos^-1(1/x)

Answer 1

\[\cos^{-1} (1/x)\]

Answer 2

Is the answer...

\[1 / (1-(1/x)^2)\]

Answer 3

Remember
\[\ \frac{ d }{ dx }\arccos x = \frac{ 1 }{ \sqrt{(1-x ^{2}} }\],
and don't forget the chain rule.

Answer 4

How do I use the chain rule for this problem?

Answer 5

\[1/\sqrt{1-(1/x)^2(2(1/x)(-1)}\]

Answer 6

Anyone?

Answer 7

let u=1/x-->u'=?
y=cos^-1(u)

Answer 8

the chain rule gives\[D_xy=\frac1{\sqrt{1-u^2}}\cdot u'\]

Answer 9

there is no "u" aside from the fact that I made it up to illustrate where the chain rule comes in

Answer 10

Oh, i see.

\[1/\sqrt{1-(1/x)^2} * -1\]

Answer 11

\[\frac d{dx}(\frac1x)\neq-1\]

Answer 12

\[\frac d{dx}(\frac1x)=\frac d{dx}(x^{-1})=?\]

Answer 13

-1(x^-2)

Answer 14

yes, that is what I called u'

Answer 15

\[1/\sqrt{1-(1/x)^2} * -1(x^-2) \]

Answer 16

\[-x^-2/\sqrt{1-(1/x)^2}\]

Answer 17

yes, which can be written\[-\frac1{x^2\sqrt{1-\frac1{x^2}}}\]or a few other ways

Answer 18

But either way is fine right?

Answer 19

Thanks, I think I understand this now.