Question

help with taking the derivative of a function..

Answer 1

http://vvcap.net/db/GzrEkdj7NVOjL0bD7HBK.htp

Answer 2

for the first half i use product rule

Answer 3

how do i differentiate 4e^(-x^(2)) ?

Answer 4

oh come on you nerds

Answer 5

use the chain rule
d/dx f(g(x)) = f'(g(x))*g'(x)
f(x)=4e^x
g(x)=-x^2

Answer 6

hmmmmm

Answer 7

yes of course... the chain rule...

Answer 8

that makes no sense buddy!

Answer 9

my solution manual isnt helping much. i tried wolfram alpha and it gave me a different answer!

Answer 10

which line is confusing?
1. d/dx f(g(x)) = f'(g(x))*g'(x)
2. f(x)=4e^x
3. g(x)=-x^2
4. f(g(x))=4e^(-x^(2))

Answer 11

4

Answer 12

how do i differentiate that?

Answer 13

calculate f'(x), then substitute in g(x) for x to get f'(g(x))
f'(x)=4e^x, thus f'(g(x))=4e^g(x)=4e^(-x^2)
to find g'(x), just take a derivative; g'(x)=-2x
d/dx f(g(x)) = f'(g(x))*g'(x) = 4e^(-x^2)*-2x

Answer 14

sorry man, i dont get it.

Answer 15

i wanted to use product rule on \[4e ^{x ^{-2}}\sin(x)\]

Answer 16

wait actually

Answer 17

the image says 4e^(-x^2) sin(x)

Answer 18

that's probably why the answer in wolfram alpha is different from the book

Answer 19

how would you do it?

Answer 20

which one, 4e^(-x^2) sin(x) or 4e^(x^(-2)) sin(x)?

Answer 21

the latter

Answer 22

1. d/dx f(g(x)) = f'(g(x))*g'(x)
2. f(x)=4e^x
3. g(x)=x^(-2)
4. f(g(x))=4e^(x^(-2))
5. f'(x)=4e^x
6. f'(g(x))=4e^g(x)=4e^(x^(-2))
7. g'(x)=-2x^(-3)
8. d/dx f(g(x)) = f'(g(x))*g'(x) = 4e^(x^(-2))*-2x^(-3)

Answer 23

what are you using? just chain rule?

Answer 24

1. d/dx f(x)*g(x) = f'(x)*g(x)+f(x)*g'(x)
2. f(x)=4e^(x^(-2))
3. g(x)=sin(x)
4. g'(x)=cos(x)
5. d/dx f(x)*g(x) = 4e^(x^(-2))*-2x^(-3)*sin(x)+4e^(x^(-2))*cos(x)

*we previously found f'(x) = 4e^(x^(-2))*-2x^(-3) using the chain rule

Answer 25

basically, first use the chain rule to find d/dx 4e^(x^(-2)), then use the product rule to find 4e^(x^(-2)) * sin(x)

Answer 26

okay cool.

Answer 27

thanks a lot man. fluttering derivatives.

Answer 28

f uck ing*