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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

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)

yes exactly

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.

one thing u have to remember is d/dx of e^x is e^x and d/dx of log x is 1/x

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

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