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sall123

  • 3 years ago

Derivative of e^(lnx)^2

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  1. TransendentialPI
    • 3 years ago
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    Use the chain rule twice.

  2. sall123
    • 3 years ago
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    how do u do it

  3. TransendentialPI
    • 3 years ago
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    derivative of e^u = e^u * du/dx What is the u in this case?

  4. sall123
    • 3 years ago
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    would it be (lnx)^2

  5. TransendentialPI
    • 3 years ago
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    right. what is the derivative of u^2?

  6. sall123
    • 3 years ago
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    2u

  7. TransendentialPI
    • 3 years ago
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    right 2u * du/dx

  8. TransendentialPI
    • 3 years ago
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    2lnx * ??

  9. sall123
    • 3 years ago
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    its 1/x^2

  10. TransendentialPI
    • 3 years ago
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    just the 1/x, the square is on the "outside"

  11. sall123
    • 3 years ago
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    therefore its 2 over x

  12. TransendentialPI
    • 3 years ago
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    put it all together. e^ln x ^2 * 2ln x * 1/x and clean it up.

  13. TransendentialPI
    • 3 years ago
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    OK? Gotta go. Good luck with it.

  14. angela210793
    • 3 years ago
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    it's e^U

  15. angela210793
    • 3 years ago
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    \[=(2\log^2_{x} logx)/x\] I guess....

  16. guyc
    • 3 years ago
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    Mmmm.... not so sure. Try this: we know d/dx(e^x) = e^x So d/dx(e^(lnx)^2) = e^(lnx)^2 times d/dx(lnx)^2 = e^(lnx)^2 times 2lnx times d/dx(lnx) = e^(lnx)^2 times 2lnx times (1/x)

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