anonymous
  • anonymous
Evaluate: anti-deriviative 3/(x-4) dx will I use U substitute
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Yes.
anonymous
  • anonymous
Identify your composite functions f(a) and g(x) such that 3/(x-4) = f(g(x))
anonymous
  • anonymous
u= x-4 du= 1

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anonymous
  • anonymous
Yes.
anonymous
  • anonymous
Well, actually not quite
anonymous
  • anonymous
du = 1dx
anonymous
  • anonymous
yes I had that just forgot to type it
anonymous
  • anonymous
ok good.
anonymous
  • anonymous
ok this is where I have trouble.....will it be 3/u ???? Doesn't look right
anonymous
  • anonymous
That is exactly right. If u = x-4 then 3/(x-4) = 3/u
anonymous
  • anonymous
so will I add 1 to the u (exponent) to be u^2
anonymous
  • anonymous
Not quite. realize that you basically have \[3 * \frac{1}{u}\] So the derivative of what function of u gives you \(\frac{1}{u}\)?
anonymous
  • anonymous
u^-1 ??
anonymous
  • anonymous
1/u is u^-1
anonymous
  • anonymous
So it's not that. \(\frac{d}{du}[ln\ u] = \frac{1}{u}\) remember?
anonymous
  • anonymous
not sure what you mean about that?
anonymous
  • anonymous
The derivative of ln x = 1/x right?
anonymous
  • anonymous
ln is the natural logarithm function.
anonymous
  • anonymous
oh, yes it does. Too many rules.....
anonymous
  • anonymous
yeah, you'll get them with practice.
anonymous
  • anonymous
So the anti derivative of 1/x = ln x + C
anonymous
  • anonymous
So 3 times the anti derivative of 1/u du = ?
anonymous
  • anonymous
ok lost again..... I know we have 3/u or 3 * 1/u would it be 3lnx + c ???
anonymous
  • anonymous
It would be 3(ln u) + C, but then you have to plug in what u is.
anonymous
  • anonymous
how about 3 ln\[\left| x-4 \right|\] + C sorry couln't get them all in one line
anonymous
  • anonymous
Yep, that's it exactly.
anonymous
  • anonymous
I am getting there......slowly....... Thanks I really appreciate your help.
anonymous
  • anonymous
Of course! Just keep practicing. You're doing very well
anonymous
  • anonymous
I have a test next Tue. So I will be asking questions up until then. I am sure I will talk to you again..
anonymous
  • anonymous
ok!
anonymous
  • anonymous
I got really good with the derivatives, so my brain still wants to do them.
anonymous
  • anonymous
Yeah, I know the feeling. Soon you'll get the hang of going the other way.
anonymous
  • anonymous
I will start one more question before I have to leave
anonymous
  • anonymous
ok

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