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ttailyn

  • 4 years ago

Find the derivative of: y=e^sec-1x (inverse trig functions)

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  1. y2o2
    • 4 years ago
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    \[y = e^{\sec^{-1}(x)} \]

  2. ttailyn
    • 4 years ago
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    yea thats how it looks.

  3. y2o2
    • 4 years ago
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    \[{e^{\sec^{-1}(x)}} \over {\sqrt{1-{1\over x^2}} \ x^2}\]

  4. y2o2
    • 4 years ago
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    http://www.wolframalpha.com/input/?i=deriv%28e%5EarcSec%28x%29%2C+x%29

  5. ttailyn
    • 4 years ago
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    thanks so much. this really helped! :)

  6. y2o2
    • 4 years ago
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    You're welcome :)

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