Question

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

Answer 1

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

Answer 2

ok i can help u

Answer 3

do u need the method or just theanswer

Answer 4

the method please

Answer 5

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)

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

what theorem or rules apply to this type of problem?

Answer 12

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

Answer 13

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)

Answer 14

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)

Answer 15

yes exactly

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

yes,that's ll

Answer 20

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

Answer 21

If you can just try to step back and see the overall structure of the question (ignoring the specific bits), sometimes that helps

Answer 22

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

Answer 23

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))

Answer 24

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

Answer 25

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. :)

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

anyways gud luck for ur exam.

Answer 30

haha thanks for everything!

Answer 31

Yes, good luck!

Answer 32

thank you