Mertsj
  • Mertsj
Derivative question
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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myininaya
  • myininaya
yes?
Mertsj
  • Mertsj
|dw:1328147554735:dw|
anonymous
  • anonymous
ho ho ho

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Mertsj
  • Mertsj
Hi Santa
myininaya
  • myininaya
Let F be the antiderivative of \[f(x)=\frac{u^2-1}{u^2+1}\] Now we have \[[g(x)]'=[F(x)|_{2x}^{3x}]=[F(3x)-F(2x)]'=(3x)'f(3x)-(2x)'f(2x)\]
myininaya
  • myininaya
that chain rule stuff is the best! :)
myininaya
  • myininaya
so we have \[3 \cdot \frac{(3x)^2-1}{(3x)^2+1}-2 \cdot \frac{(2x)^2-1}{(2x)^2+1}\]
myininaya
  • myininaya
\[3 \cdot \frac{9x^2-1}{9x^2+1}-2 \cdot \frac{4x^2-1}{4x^2+1}\]
anonymous
  • anonymous
\[-\int _a^{2x}\frac{u^2-1}{u^2+1}du+\int_a^{3x}\frac{u^2-1}{u^2+1}du\]
anonymous
  • anonymous
dang did that take me a long time to type now chain rule
myininaya
  • myininaya
yes it did satellite :)
anonymous
  • anonymous
i am tired, time for bed. gnight
myininaya
  • myininaya
me too we are going to bed
Mertsj
  • Mertsj
\[g'(x)=\frac{-2(4x^2-1)}{4x^2+1}+\frac{3(9x^2-1)}{9x^2+1}\]
Mertsj
  • Mertsj
Not yet, I hope.
Mertsj
  • Mertsj
I don't get the -2 and the 3
Mertsj
  • Mertsj
Maybe if I study it.
myininaya
  • myininaya
:do you understand what I did?
Mertsj
  • Mertsj
Not quite.
Mertsj
  • Mertsj
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.
myininaya
  • myininaya
\[g(x)=\int\limits_{2x}^{3x}f(u) du=F(u)|_{2x}^{3x}=F(3x)-F(2x)\] Do you understand this part?
Mertsj
  • Mertsj
Why isn't f(u) the g'(x)?
myininaya
  • myininaya
no \[g(x)=F(3x)-F(2x)\]
myininaya
  • myininaya
g is the integral from 2x to 3x of f(u) with respect to u
myininaya
  • myininaya
now since we know that \[g(x)=F(3x)-F(2x)\] we can differentiate both sides
Mertsj
  • Mertsj
Ok. I get that and I understand what you wrote about F(3x)-F(2x)
myininaya
  • myininaya
\[g'(x)=(3x)'f(3x)-(2x)'f(2x)\]
myininaya
  • myininaya
i used chain rule for this
Mertsj
  • Mertsj
One second.
myininaya
  • myininaya
\[g'(x)=3 f(3x)-2 f(2x)\] since (3x)'=3 and (2x)'=2
myininaya
  • myininaya
ok one sec granted
Mertsj
  • Mertsj
so F(3x)=(3x)'f(x)?
Mertsj
  • Mertsj
And what happened to F(x)?
myininaya
  • myininaya
well (F(3x))'=(3x)'f(3x) by chain rule
myininaya
  • myininaya
remember we differentiate both sides so we could find g'
Mertsj
  • Mertsj
Yep. Ok. I think i can figure it out now. thank you very much and how did you get so smart?
myininaya
  • myininaya
i'm a little smart not much
myininaya
  • myininaya
well i'm not really smart lol
myininaya
  • myininaya
but thanks mert
Mertsj
  • Mertsj
thanks to you and sleep well...
myininaya
  • myininaya
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
myininaya
  • myininaya
he likes to split the integral at a constant
myininaya
  • myininaya
and then use the other part of the fundamental thm of calculus
myininaya
  • myininaya
goodnight! :)
Mertsj
  • Mertsj
Satellite is very smart also. Do you know him personally?

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