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101Ryan101

  • 3 years ago

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

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  1. Diogo
    • 3 years ago
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    arcsec(x) = 1/(sqrt(1-1/x^2) x^2)

  2. malevolence19
    • 3 years ago
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    \[\frac{d}{dx} \sec^{-1}(x)=\frac{1}{|x|\sqrt{x^2-1}}\] You can drop the ||'s if x is positive.

  3. myininaya
    • 3 years ago
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    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
    • 3 years ago
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    by the way tan(y) is opposite/adjacent

  5. 101Ryan101
    • 3 years ago
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    very funny

  6. myininaya
    • 3 years ago
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    whats funny lol

  7. 101Ryan101
    • 3 years ago
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    thanks myininaya

  8. myininaya
    • 3 years ago
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    np

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