Question

Differentiation:

Answer 1

Let \(y=sec^{-1}\) x. Find \(y'=\frac{d}{dx}~sec^{-1}~x\).

Answer 2

first
y=cosx
right

Answer 3

ok
secy =x
right?

Answer 4

then
secy tany *dy/dx=1

Answer 5

y = sec^-1 x

sec y = x

1/cos y = x

cos y = 1/x

y = cos^-1 (1/x)

Answer 6

How is y = cos x? O.O

Answer 7

rule
costheta=1/sectheta

Answer 8

Yes, I am aware of that.

But this is an inverse function whose derivative has to be found.
It is not saying (sec^-1 x = 1/cos x)

Answer 9

ok sorry :)

Answer 10

I am not sure how to do this question, truth be told. :)

@Luigi0210, you should ask @satellite73 or @ganeshie.

Answer 11

Thanks for trying >.<
We're suppose to use a right triangle to go along with this .-.

Answer 12

@whpalmer4 @jdoe0001

Answer 13

First, reverse it: \(\sf x =\color{red}{arcec(y)}\)
now, differentiate with respect to \(x\)

Answer 14

Can you do this now?

Answer 15

inverse it* and i mean't arcsec, not arcec. Lol
also NOTE: arcec = sec\(^{-1}\)(x)

Answer 16

arsec*

Answer 17

ugh. ARCSEC* lol

Answer 18

noob ^

Answer 19

Woah, calm down guys, things are heating up here.