anonymous
  • anonymous
whats the derivative to arcsecx? or sec^(-1)x
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
arcsec(x) = 1/(sqrt(1-1/x^2) x^2)
anonymous
  • anonymous
\[\frac{d}{dx} \sec^{-1}(x)=\frac{1}{|x|\sqrt{x^2-1}}\] You can drop the ||'s if x is positive.
myininaya
  • myininaya
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 \[y'=\frac{1}{x*\frac{\sqrt{x^2-1}}{1}}=\frac{1}{x \sqrt{x^2-1}}\]

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myininaya
  • myininaya
by the way tan(y) is opposite/adjacent
anonymous
  • anonymous
very funny
myininaya
  • myininaya
whats funny lol
anonymous
  • anonymous
thanks myininaya
myininaya
  • myininaya
np

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