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Derivative question

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ho ho ho

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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
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)
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?

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