Question

Find the first and second derivitaives of the function
y= e^e^x

Answer 1

use chain rule

Answer 2

omg taht is sooo confusing wel atleast how my teacher explains it

Answer 3

y'=e^e^x.e^x

Answer 4

is taht a multiplication symbol btwn e^e^x and e^x?

Answer 5

yes

Answer 6

\[y=e ^{e ^{x}}\]

Answer 7

do u know chain rule?

Answer 8

Just break it down, derivative of e^x would be e^x, which is the exponent for the lower e...therefore you would multiple e^e^x by e^x.

Answer 9

i knw teh chain rule but idk how 2 use it

Answer 10

For the second derivative you need to chain rule and product rule

Answer 11

i know teh product rule

Answer 12

so im confused about this chain rule

Answer 13

y'=e^e^x.e^x u understand this?

Answer 14

now product rule

Answer 15

: f'(x) g(x) is taht wat u did?

Answer 16

(fg)'=f'g+fg'

Answer 17

Uzma please correct me if I made a mistake but the second derivative is y''= \[e^x *e^e^x + (e^x)^2 * e^e^x \]

Answer 18

u r right jm

Answer 19

so the anwer to the frst der is e^e^x * e^x
and teh 2nd der is wat is written above?

Answer 20

Yes.

Answer 21

yea

Answer 22

but make sure that u understand it

Answer 23

can u explain teh chain rule

Answer 24

if u have y=f(x)^g(x)
the y'=[f(x)^g(x)]'.[g(x)]'

Answer 25

in this case it was f(x) ^g(x) to another exponent....so then how wud u modify the formula

Answer 26

no modification, just classify f(x) and g(x)

Answer 27

wat do u mean?

Answer 28

f(x)=e^e^x and g(x)=e^x
apply the rule

Answer 29

ohhh i get it.....

Answer 30

so the der of e^e^x is just e^e^x? since e^x is e^x

Answer 31

\[y[x]\text{=}e^{e^x} \]\[y'[x]=e^{e^x+x} \text{Log}[e]^2 \]\[y\text{''}[x]=e^{e^x+x} \text{Log}[e]^3 \left(1+e^x \text{Log}[e]\right) \]

Answer 32

whoah...now u introduced logs..im confused....

Answer 33

der of e^e^x is e^e^x, right?

Answer 34

yes

Answer 35

the what is exponent?
e^x

Answer 36

and its der ise^x
multiply both

Answer 37

inst it just e^x^2

Answer 38

no

Answer 39

idk