Question

Calculus @mathmale @zepdrix

Answer 1

\[\large\rm f(x)=\int\limits_0^{x^3} (t^2+4)dt\]

Answer 2

Do I plug in x^2 for t?

Answer 3

Remember your fundamental theorem?\[\large\rm \frac{d}{dx}\int\limits_c^x f(t)dt=f(x)\]

Answer 4

You will plug in x^3 for t, yes,
but you will also chain rule, because the upper bound on your limit `is more than just x`

Answer 5

So, mI also multiply by the derivative of x^2? Which is 2x?

Answer 6

I*

Answer 7

Here is an extension of the FTC, which applies more generally to a problem like this:\[\large\rm \frac{d}{dx}\int\limits_c^{g(x)}f(t)dt\quad=\quad f(g(x))\cdot g'(x)\]

Answer 8

Am I reading the problem wrong?
Isn't x^3, isn't it?

Answer 9

yep, sorry, my mistake. x^3

Answer 10

so i\(\Large f(x) = (x^6 + 4)3x^2=3x^8 + 12x^2\)?

Answer 11

\[\large\rm \frac{d}{dx}\int\limits\limits_c^{\color{orangered}{g(x)}}f(t)dt\quad=\quad f(\color{orangered}{g(x)})\cdot \frac{d}{dx}\color{orangered}{g(x)}\]So we have,\[\large\rm \frac{d}{dx}\int\limits\limits_c^{\color{orangered}{x^3}}f(t)dt\quad=\quad f(\color{orangered}{x^3})\cdot \frac{d}{dx}\color{orangered}{x^3}\]Ooo yes, good job! :)

Answer 12

So then I just find the inverse of that?

Answer 13

No, you're done :o

Answer 14

But it says find f'(x)

Answer 15

Oh, wait, sorry. Not the inverse, I meant the dderivative

Answer 16

@zepdrix ?

Answer 17

\[\large\rm f(x)=\int\limits\limits_0^{x^3} (t^2+4)dt\]So then,\[\large\rm \frac{d}{dx}f(x)=\frac{d}{dx}\int\limits\limits_0^{x^3} (t^2+4)dt\]And by the Fundamental Theorem, we found that the right side becomes,\[\large\rm \frac{d}{dx}f(x)=3x^8+12x^2\]

Answer 18

So we have found our derivative f'(x), yes?

Answer 19

I don't get it...

Answer 20

hmm

Answer 21

\[\large\rm \frac{d}{dx}f(x)=f'(x)\]Yes?
We took derivative of each side.

Answer 22

so basically you're saying that f'(x) = f(x)

Answer 23

Oh oh oh,
I shouldn't have used f in my definition :(
Maybe that's what's causing the confusion.
\[\large\rm \frac{d}{dx}\int\limits\limits\limits_c^{\color{orangered}{g(x)}}h(t)dt\quad=\quad h(\color{orangered}{g(x)})\cdot \frac{d}{dx}\color{orangered}{g(x)}\]So we started with this,\[\large\rm f(x)=\int\limits\limits\limits_0^{x^3} (t^2+4)dt\]Which we can think of more generally as this,\[\large\rm f(x)=\int\limits\limits\limits_0^{x^3} h(t)dt\]Taking derivative of each side,\[\large\rm \frac{d}{dx}f(x)=\frac{d}{dx}\int\limits\limits\limits_0^{x^3} h(t)dt\]so then by FTC,\[\large\rm f'(x)=h(x^3)\cdot(x^3)'\]

Answer 24

I dunno if that helps or not :d
Just wanted to clear up the discrepancy with the function labeling.

Answer 25

because the derivative of the integral cancels out, and what you get is basically h(t), plus whatever chain rule stuff apples?

Answer 26

Yes, they "sort of" cancel out.
You're taking an anti-derivative, then you're taking derivative,
so you get back what you started with.

But something happens `between those two steps`.
You're plugging in limits, so the argument of your function changes.

Answer 27

I see. Thank you!

Answer 28

cool c:
this one can be a tricky rule to get a hold of!
practice a few more of them!
especially ones like this:\[\large\rm f(x)=\int\limits_{x^2}^{x^3}\sqrt{t^2+3}~dt\]Where the upper AND lower bound contain x stuff.