At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

See more answers at brainly.com

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the **expert** answer you'll need to create a **free** account at **Brainly**

yes?

|dw:1328147554735:dw|

ho ho ho

Hi Santa

that chain rule stuff is the best! :)

so we have
\[3 \cdot \frac{(3x)^2-1}{(3x)^2+1}-2 \cdot \frac{(2x)^2-1}{(2x)^2+1}\]

\[3 \cdot \frac{9x^2-1}{9x^2+1}-2 \cdot \frac{4x^2-1}{4x^2+1}\]

\[-\int _a^{2x}\frac{u^2-1}{u^2+1}du+\int_a^{3x}\frac{u^2-1}{u^2+1}du\]

dang did that take me a long time to type
now chain rule

yes it did satellite :)

i am tired, time for bed. gnight

me too
we are going to bed

\[g'(x)=\frac{-2(4x^2-1)}{4x^2+1}+\frac{3(9x^2-1)}{9x^2+1}\]

Not yet, I hope.

I don't get the -2 and the 3

Maybe if I study it.

:do you understand what I did?

Not quite.

\[g(x)=\int\limits_{2x}^{3x}f(u) du=F(u)|_{2x}^{3x}=F(3x)-F(2x)\]
Do you understand this part?

Why isn't f(u) the g'(x)?

no \[g(x)=F(3x)-F(2x)\]

g is the integral from 2x to 3x of f(u) with respect to u

now since we know that
\[g(x)=F(3x)-F(2x)\]
we can differentiate both sides

Ok. I get that and I understand what you wrote about F(3x)-F(2x)

\[g'(x)=(3x)'f(3x)-(2x)'f(2x)\]

i used chain rule for this

One second.

\[g'(x)=3 f(3x)-2 f(2x)\]
since (3x)'=3 and (2x)'=2

ok one sec granted

so F(3x)=(3x)'f(x)?

And what happened to F(x)?

well (F(3x))'=(3x)'f(3x) by chain rule

remember we differentiate both sides so we could find g'

Yep. Ok. I think i can figure it out now. thank you very much and how did you get so smart?

i'm a little smart not much

well i'm not really smart lol

but thanks mert

thanks to you and sleep well...

he likes to split the integral at a constant

and then use the other part of the fundamental thm of calculus

goodnight! :)

Satellite is very smart also. Do you know him personally?