Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

Derivative question

Mathematics
See more answers at brainly.com
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.

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Hi Santa
Let F be the antiderivative of \[f(x)=\frac{u^2-1}{u^2+1}\] Now we have \[[g(x)]'=[F(x)|_{2x}^{3x}]=[F(3x)-F(2x)]'=(3x)'f(3x)-(2x)'f(2x)\]
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.
I thought the u^2-1/u^2+1 would be the derivative and all I had to do was plug in the interval endpoints.
\[g(x)=\int\limits_{2x}^{3x}f(u) du=F(u)|_{2x}^{3x}=F(3x)-F(2x)\] 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...
thanks i'm going to sleep for reals i'm so tired by the way satellite likes to do these types of problems another way but i love my way
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?

Not the answer you are looking for?

Search for more explanations.

Ask your own question