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steel11

  • 3 years ago

differentiate ln(7e^-x+xe^-x)

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  1. steel11
    • 3 years ago
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    my answer was 1/f(x) * f'(x). but it keeps saying wrong

  2. hartnn
    • 3 years ago
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    first take e^(-x) common from (7e^-x+xe^-x) then use the property of log. log ab=log a + log b

  3. steel11
    • 3 years ago
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    so, ln e^-x(7+x)?

  4. cinar
    • 3 years ago
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    answer=(-7e^-x+e^-x-xe^-x)/(7e^-x+xe^-x)

  5. hartnn
    • 3 years ago
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    ln [e^-x(7+x)] now use log ab = log a +log b

  6. steel11
    • 3 years ago
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    so, lne^-x+ln(7+x), and go from there?

  7. hartnn
    • 3 years ago
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    u can simplify ln(e^(-x)) do u know how ?

  8. steel11
    • 3 years ago
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    no, im afraid not goku. :l

  9. hartnn
    • 3 years ago
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    \(\huge ln e^{(a)}=aln e=a\) now ?

  10. steel11
    • 3 years ago
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    oh ye, makes sense.

  11. hartnn
    • 3 years ago
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    so lne^-x+ln(7+x) = ?

  12. steel11
    • 3 years ago
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    so, -x+ln(7+x)? is that right so far?

  13. hartnn
    • 3 years ago
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    yes, that is correct :) now its differentiation is simple....

  14. steel11
    • 3 years ago
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    -1+...ermm 1/(7+x)*1?

  15. hartnn
    • 3 years ago
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    that is absolutely correct (except for ...ermm part) simplify it further

  16. steel11
    • 3 years ago
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    lol, wow. thank you goku san.

  17. hartnn
    • 3 years ago
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    welcome ^_^

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