Question

what am i doing wrong here ?

Let "picture" Use your calculator to find F″(1)

Answer 1

im getting 2

Answer 2

but these are my only answer choices
12
6
4
1/9

Answer 3

F'(x) = ln(9x^2)
F"(x) = 2/x
F"(1) = 2
what am i doing wrong?

Answer 4

@abb0t @freckles @UsukiDoll

Answer 5

~_O uhhhhh.. integration is always anti-derivatives no matter what... something is off.

Answer 6

do you hink the answer choices are somehow wrong?

Answer 7

ah... try taking the anti-derivative.... then f(b)-f(a) which is f(3x)- f(1) and then evaluate when x = 1?!?!?!!?!?!!!!!!!!!!!!!!!!!!!!!!!!!!

this is one strange question

Answer 8

because whenever I see the integral sign... it's always antiderivative.

Answer 9

@ganeshie8 any ideas? O_o

Answer 10

hey @Astrophysics while i have you here can you give me the link to the cool openstudy extension i managed to lose the question where you provided the links :/

Answer 11

Hey sorry I was away, yeah sure just send me a message of what exactly you want.

since F'(x) = f(x) we have\[F'(x) = \int\limits_{1}^{3x} \ln(t^2) dt\]

\[F'(x) = \ln(9x^2) - \ln(1)\]

F''(x) = f'(x) \[F''(x) = \frac{ 2 }{ x }\]

\[F''(1) = 2\] so this is what you did?

Answer 12

yes

Answer 13

Mhm that seems right to me haha.

Answer 14

so somehow the answer choices and whatever you guys did don't match... :/ ~_O! either wrong problem screenshot or wrong answer choices?!!!!!

Answer 15

\[\frac{ d }{ dx } \int\limits_{1}^{3x} \ln(t^2) dt = \ln(3x)\] see it basically just cancels out the integral you get what you initially started with, but I don't really trust this integral itself...

Answer 16

I got the solution.. *grabs problems and throws it away*

Answer 17

I think there is a reason it asks us to use a calculator :P

Answer 18

the ln(1) = 0

Answer 19

yeah I was thinking that but it truly no longer matters i submitted the answer as 4 and got it wrong , no worries though ill just ask my professor how to do it and see what happens.

Answer 20

uh huh ln 1 is 0...
the way the question is written is odd...too strange for me :P

Answer 21

fundamental theorem of calculus? eh forget it........ <_<

Answer 22

Ah wait let me see...F'(x) = f(x) \[F'(x) = \int\limits_{0}^{3x} \ln(t^2) dt = \ln(9x^2)-\ln(1) \]

\[F'(x) = \ln(9x^2) \times (3x)' = 3\ln(9x^2) \] I found the mistake...we're suppose to take the derivative of 3x as well..totally missed that part...ok so we have now

F''(x) = f'(x)

\[F''(x) = \frac{ 6 }{ x } \implies f''(1)= 6\]

Answer 23

F''(1) = 6*

Answer 24

Fundamental theorem of calculus..bleh

Answer 25

AAARRRRGGGGHHHH

Answer 26

wait wtf so when I said Fundamental theorem of Calculus, we got wtf we need?

Answer 27

Yeah haha...can you submit your answer again, does it show the right one?

Answer 28

da phuc?

Answer 29

haha

Answer 30

no -_- but it is clearly 6

Answer 31

damn it all to hell D:

Answer 32

ffffffffffuuuuuuxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Answer 33

ugh oh well , got a B either way

Answer 34

I apologize for the late reply, sorry @Jdosio :\

Answer 35

its okay, thats what i get when i don't wait for you :(

Answer 36

ah I see how FTC part 1 would work... but damn too late...

Answer 37

Yeah haha, I just remembered part 1, it's so silly

Answer 38

then product rule twice... -_- CURSESSSSSSSSSSSSSSSSSSSSSS

Answer 39

thanks for sticking around so long either way :)

Answer 40

the variable on f(b) I remember using FTC part 1 all the time on that one... you have to substitute it into the equation (that ln t^2) and then take the derivative of 3x as well

Answer 41

Yup

Answer 42

You have to use the chain rule pretty much...it's weird.

Answer 43

phuc.. it was my slow thinking -_-

Answer 44

Maybe you can get your mark back, just show the work I guess

Answer 45

To your professor

Answer 46

its fine its just one question,

Answer 47

Well you won't forget next time either, as I like to say...the struggle is more important than the final answer :)