Question

Find F'(x) if F(x) = integral from 1 to 3x^2 of 1/t dt.

Answer 1

F(x) = \[\int\limits_{1}^{3x^2} \frac{ 1 }{ t } dt\]

Answer 2

Fundamental theorem of calculus?

Answer 3

I know the fundemental theorum of calculus applies.
But I don't know how to use it here.
I would just say 1/3x^2 - 1 but I'm pretty sure that's wrong. ( I never understood how to do this.)

Answer 4

Set \(y = 3x^2\) and remember that: \[
\frac{dF}{dx} = \frac{dF}{dy}\frac{dy}{dx}
\]

Answer 5

wait. i think i got it. is it F'(x) = 12/x ?

Answer 6

sorry for no work. :(
is it just take the derivative and plug it in?

Answer 7

First of all, what is \(dy/dx\)?

Answer 8

6x

Answer 9

What's \(dF/dy\)?

Answer 10

AHHHH sheet. I typed it wrong...again.
It's supposed to be
F(x) = \[\int\limits_{1}^{3x^12} \frac{ 1 }{ t }dt\]

Answer 11

isn't dF/dy just 1/t ? i'll stick with my old question

Answer 12

It's \(1/y = 1/(3x^2)\)

Answer 13

wait what? i thought it was derviatvie of F with respect to y...?

Answer 14

oh yeah sorry i see what u did.

Answer 15

\[
\frac{d}{dy}\int^y_af(t)dt = f(y)=f(y(x))
\]

Answer 16

Forgetting everything, let's just do the integral?
\[\huge \int\limits\limits_{1}^{3x^{12}} \frac{ 1 }{ t }dt = \ln(3x^{12})-\ln(1)=\ln(3x^{12})\]
And then get its derivative?

Answer 17

oh ok i get it now...
so it's just (36x^11) / ln (3x^12) then ?

Answer 18

Why is there still a logarithm?

Answer 19

oh wait whoops. forgot that. my computer's slow so i thought i had already erased that.

Answer 20

yeah it's 36x^11/3x^12 which simplifes to 12/x right?

Answer 21

It should be
\[\huge \frac{36x^{11}}{3x^{12}}=\frac{12}{x}\]
Or, cleverly,
\[\large \ln(3x^{12})=\ln(3) + \ln(x^{12})\]
And then get the derivative

Either way, you're right :)

Answer 22

oh okay thanks.
i kept not understanding that up to now.

Answer 23

:)