Question

Calculus Help only 1 Question.

Answer 1

\[F(x) = \int_a^x f(s) ds\] where \[F'(x) = f(x).\]

Answer 2

@jhonyy9 @dan815 @DanJS @pooja195 @Kkutie7 @katherinkoon

Answer 3

Well I know they are asking for the derivative....

Answer 4

I hate that they used x in the limits, and x in the function that you're integrating,
so sloppy D:

Anyway, umm

Answer 5

This is actually the First Fundamental Theorem of Calculus,
I guess your class has them listed the other way though.
\[\large\rm \frac{d}{dx}\int\limits_a^{2x} f(s)ds\]So let's say that f(s) has some anti-derivative that we will decide to call F(s).\[\large\rm \frac{d}{dx}\int\limits\limits_a^{2x} f(s)ds\quad=\quad \frac{d}{dx}F(s)|_a^{2x}\]Using our other fundamental theorem,
we can evaluate this at the limits,\[\large\rm =\frac{d}{dx}\left[F(2x)-F(a)\right]\]Now we take derivative,
big F turns back into little f,
We have chain rule in the first term,
the second term is a constant, so it's derivative is 0.\[\large\rm = f(2x)(2x)'-0\]

Answer 6

So for your problem, you're starting with something like this,\[\large\rm F(x)=\int\limits_0^{2x}\frac{1}{t^2}dt=\int\limits_0^{2x} f(t)dt\]The idea behind this process is that we're finding an anti-derivative,
and then the derivative of that,
so we end up with the thing we started with,
just in terms of x instead of our t, whatever it was before.

Answer 7

\[\large\rm F'(x)=f(2x)\cdot (2x)'\]

Answer 8

\[\large\rm F'(x)=\frac{1}{(2x)^2}(2x)'\]

Answer 9

Too confusing? D:

Answer 10

Nope I think I get it.

Answer 11

but u used first... how do i explain using it second?

Answer 12

One of our FTC's tells us that,\[\large\rm \frac{d}{dx}\int\limits_a^x f(t)dt=f(x)\]
I was using the "other" FTC to put some detail into why this is true,\[\large\rm \frac{d}{dx}\int\limits_0^{2x}f(t)dt=f(2x)\cdot(2x)'\]

Answer 13

I guess you should look at your FTC more generally like this,
to match our problem a little better,
it might look confusing though,\[\large\rm \frac{d}{dx}\int\limits_0^{g(x)}f(t)dt=f(g(x))\cdot g'(x)\]

Answer 14

It's the same function we started with,
but with g(x) plugged in place of the t,
and chain rule.

Answer 15

But, yes, you're correct,
we don't NEED to use the other FTC, that was just to give some details :)

Answer 16

Using only the one FTC rule,\[\rm \frac{d}{dx}\int\limits\limits_0^{2x}\frac{1}{t^2}dt=\frac{d}{dx}\int\limits\limits_0^{2x}f(t)dt=f(2x)(2x)'=\frac{1}{(2x)^2}(2x)'=\frac{1}{(2x)^2}(2)\]