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## 101Ryan101 3 years ago whats the derivative to arcsecx? or sec^(-1)x

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1. Diogo

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

2. malevolence19

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

3. 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 $y'=\frac{1}{x*\frac{\sqrt{x^2-1}}{1}}=\frac{1}{x \sqrt{x^2-1}}$

4. myininaya

by the way tan(y) is opposite/adjacent

5. 101Ryan101

very funny

6. myininaya

whats funny lol

7. 101Ryan101

thanks myininaya

8. myininaya

np

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