Question

whats the derivative to arcsecx? or sec^(-1)x

Answer 1

arcsec(x) = 1/(sqrt(1-1/x^2) x^2)

Answer 2

\[\frac{d}{dx} \sec^{-1}(x)=\frac{1}{|x|\sqrt{x^2-1}}\]
You can drop the ||'s if x is positive.

Answer 3

let \[y=\sec^{-1}(x)\]
so \[\sec(y)=x\]
therefore \[y'\sec(y)\tan(y)=1\]
=> \[y'=\frac{1}{\sec(y)\tan(y)}\]
but we need this in terms of x
sec(y)=x remember!
and we can find tan(y) by looking at what sec(y) means
sec(y)=hyp/adj=x/1

so the missing side is opposite to y so we can find it by doing sqrt{x^2-1}

so we have
\[y'=\frac{1}{x*\frac{\sqrt{x^2-1}}{1}}=\frac{1}{x \sqrt{x^2-1}}\]

Answer 4

by the way tan(y) is opposite/adjacent

Answer 5

very funny

Answer 6

whats funny lol

Answer 7

thanks myininaya

Answer 8

np