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anonymous

  • 5 years ago

how do i find the derivative of f(x) = e^(1-2x^2) + ln (3-e^-4x)?

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  1. anonymous
    • 5 years ago
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    i need the answer...my exam is tomorrow! please help :(

  2. anonymous
    • 5 years ago
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    ok i can help u

  3. anonymous
    • 5 years ago
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    do u need the method or just theanswer

  4. anonymous
    • 5 years ago
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    the method please

  5. anonymous
    • 5 years ago
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    ok here it goes d/dx(e^(1-2x^2)) is -4xe^(1-2x^2) because d/dx(e^f(x)) is e^(x)*d/dx(f(x)

  6. anonymous
    • 5 years ago
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    here f(x) is 1-2x^2 therefore d/dx(1-2x^2) is -4x

  7. anonymous
    • 5 years ago
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    then d/dx(ln(3-e^-4x) is 4e^-4x/(3-e^-4x)

  8. anonymous
    • 5 years ago
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    because d/dx(log(f(x)) is (1/f(x))*d/dxf(x)

  9. anonymous
    • 5 years ago
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    where f(x) is (3-e^-4x)

  10. anonymous
    • 5 years ago
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    so the answer is -4xe^(1-2x^2)+4e^-4x/(3-e^-4x)

  11. anonymous
    • 5 years ago
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    what theorem or rules apply to this type of problem?

  12. anonymous
    • 5 years ago
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    first of all d/dx(a+b) is da/dx+db/dx where a is e^(1-2x^2) and b is ln(3-e^-4x

  13. anonymous
    • 5 years ago
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    first of all d/dx(a+b) is da/dx+db/dx where a is e^(1-2x^2) and b is ln(3-e^-4x)

  14. anonymous
    • 5 years ago
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    Maybe this will help: the derivative of e ^ (stuff) = e ^ (stuff) * (derivative of that stuff) the derivative of ln (stuff) = 1/(stuff) * (derivative of that stuff)

  15. anonymous
    • 5 years ago
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    yes exactly

  16. anonymous
    • 5 years ago
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    And as dipin said, if you have the derivative of things added or subtracted, you just take each derivative as you come across it. So derivative of (A + B) = derivative of A + derivative of B

  17. anonymous
    • 5 years ago
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    It's just a big ugly when you have to type these things instead of write them by hand. :)

  18. anonymous
    • 5 years ago
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    so we just solve the e^stuff and ln^stuff and then add them?

  19. anonymous
    • 5 years ago
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    yes,that's ll

  20. anonymous
    • 5 years ago
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    i just get confused because of the logs, e's, and ln's.

  21. anonymous
    • 5 years ago
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    If you can just try to step back and see the overall structure of the question (ignoring the specific bits), sometimes that helps

  22. anonymous
    • 5 years ago
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    one thing u have to remember is d/dx of e^x is e^x and d/dx of log x is 1/x

  23. anonymous
    • 5 years ago
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    Once you know the derivative of e^(stuff) = e^(stuff) * (derivative of that stuff), then you can make the stuff as complicated as you want, and you still have a basic framework for solving the question. (Of course, assuming all the proper variables like the derivative we're taking is dy/dx and the function is expressed in terms of x (or x and y, if you're there yet))

  24. anonymous
    • 5 years ago
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    for example e^4x is e^4x*d/dx(4x) which is 4e^4x

  25. anonymous
    • 5 years ago
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    You don't even have to remember d/dx e^x = e^x, if you don't want to, because if you follow the pattern, you'll get there anyway. e^x just the special case where the (stuff) is x. And the derivative of x = 1. So, you get e^x * 1 when you take the derivative with respect to x. That's the nice thing about following the rules. :)

  26. anonymous
    • 5 years ago
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    And then, you just clean it up like dipin does, usually writing it cleanly and compactly

  27. anonymous
    • 5 years ago
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    d/dx(1-2x^2) will be -2*2*x which is -4x and we have a e^(stuff) there

  28. anonymous
    • 5 years ago
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    oh! can i still stuff you guys into my head during the exam? :)

  29. anonymous
    • 5 years ago
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    anyways gud luck for ur exam.

  30. anonymous
    • 5 years ago
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    haha thanks for everything!

  31. anonymous
    • 5 years ago
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    Yes, good luck!

  32. anonymous
    • 5 years ago
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    thank you

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