anonymous
  • anonymous
how do i find the derivative of f(x) = e^(1-2x^2) + ln (3-e^-4x)?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
i need the answer...my exam is tomorrow! please help :(
anonymous
  • anonymous
ok i can help u
anonymous
  • anonymous
do u need the method or just theanswer

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anonymous
  • anonymous
the method please
anonymous
  • anonymous
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)
anonymous
  • anonymous
here f(x) is 1-2x^2 therefore d/dx(1-2x^2) is -4x
anonymous
  • anonymous
then d/dx(ln(3-e^-4x) is 4e^-4x/(3-e^-4x)
anonymous
  • anonymous
because d/dx(log(f(x)) is (1/f(x))*d/dxf(x)
anonymous
  • anonymous
where f(x) is (3-e^-4x)
anonymous
  • anonymous
so the answer is -4xe^(1-2x^2)+4e^-4x/(3-e^-4x)
anonymous
  • anonymous
what theorem or rules apply to this type of problem?
anonymous
  • anonymous
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
anonymous
  • anonymous
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)
anonymous
  • anonymous
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)
anonymous
  • anonymous
yes exactly
anonymous
  • anonymous
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
anonymous
  • anonymous
It's just a big ugly when you have to type these things instead of write them by hand. :)
anonymous
  • anonymous
so we just solve the e^stuff and ln^stuff and then add them?
anonymous
  • anonymous
yes,that's ll
anonymous
  • anonymous
i just get confused because of the logs, e's, and ln's.
anonymous
  • anonymous
If you can just try to step back and see the overall structure of the question (ignoring the specific bits), sometimes that helps
anonymous
  • anonymous
one thing u have to remember is d/dx of e^x is e^x and d/dx of log x is 1/x
anonymous
  • anonymous
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))
anonymous
  • anonymous
for example e^4x is e^4x*d/dx(4x) which is 4e^4x
anonymous
  • anonymous
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. :)
anonymous
  • anonymous
And then, you just clean it up like dipin does, usually writing it cleanly and compactly
anonymous
  • anonymous
d/dx(1-2x^2) will be -2*2*x which is -4x and we have a e^(stuff) there
anonymous
  • anonymous
oh! can i still stuff you guys into my head during the exam? :)
anonymous
  • anonymous
anyways gud luck for ur exam.
anonymous
  • anonymous
haha thanks for everything!
anonymous
  • anonymous
Yes, good luck!
anonymous
  • anonymous
thank you

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