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

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

Mathematics
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i need the answer...my exam is tomorrow! please help :(
ok i can help u
do u need the method or just theanswer

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Other answers:

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

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