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

- anonymous

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

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- schrodinger

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- anonymous

i need the answer...my exam is tomorrow! please help :(

- anonymous

ok i can help u

- anonymous

do u need the method or just theanswer

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## More answers

- anonymous

the method please

- 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

here f(x) is 1-2x^2 therefore d/dx(1-2x^2) is -4x

- anonymous

then d/dx(ln(3-e^-4x) is 4e^-4x/(3-e^-4x)

- anonymous

because d/dx(log(f(x)) is (1/f(x))*d/dxf(x)

- anonymous

where f(x) is (3-e^-4x)

- anonymous

so the answer is -4xe^(1-2x^2)+4e^-4x/(3-e^-4x)

- anonymous

what theorem or rules apply to this type of problem?

- 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

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

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

yes exactly

- 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

It's just a big ugly when you have to type these things instead of write them by hand. :)

- anonymous

so we just solve the e^stuff and ln^stuff and then add them?

- anonymous

yes,that's ll

- anonymous

i just get confused because of the logs, e's, and ln's.

- 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

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

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

for example e^4x is e^4x*d/dx(4x) which is 4e^4x

- 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

And then, you just clean it up like dipin does, usually writing it cleanly and compactly

- anonymous

d/dx(1-2x^2) will be -2*2*x which is -4x and we have a e^(stuff) there

- anonymous

oh! can i still stuff you guys into my head during the exam? :)

- anonymous

anyways gud luck for ur exam.

- anonymous

haha thanks for everything!

- anonymous

Yes, good luck!

- anonymous

thank you

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