anonymous
  • anonymous
As soon as I think I've finished, more work appears! YAY! (EXTREME sarcasm). I'll give a medal to anyone who can help me with the problem I'm posting below. :)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[(x^2-1)^\frac{5}{2} (x^3+5)\]
jim_thompson5910
  • jim_thompson5910
what are the instructions?
anonymous
  • anonymous
Sorry, the directions are to find the derivative. (not sure why i didn't post that?)

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anonymous
  • anonymous
So far I've gotten...
anonymous
  • anonymous
\[\:\left(x^2-1\right)^{\frac{5}{2}}\left(3x^2\right)+\left(x^3+5\right)\left(\frac{5}{2}\right)\left(x^2-1\right)^{\frac{3}{2}}\left(2x\right)\]
jim_thompson5910
  • jim_thompson5910
it looks great
anonymous
  • anonymous
I'm not exactly sure what I should do next. Should I begin to simplify or is there another step involved?
jim_thompson5910
  • jim_thompson5910
there's not much to simplify after this point
anonymous
  • anonymous
So that's it?
jim_thompson5910
  • jim_thompson5910
the only thing I can see really is factoring but it's not much of a simplification if you ask me
anonymous
  • anonymous
My teacher prefers specifics unfortunately. Could you help me identify what parts should be factored?
jim_thompson5910
  • jim_thompson5910
notice there's a x^2-1 repeated
jim_thompson5910
  • jim_thompson5910
\[\large \left(x^2-1\right)^{\frac{5}{2}}\left(3x^2\right)+\left(x^3+5\right)\left(\frac{5}{2}\right)\left(x^2-1\right)^{\frac{3}{2}}\left(2x\right)\] \[\large \color{red}{\left(x^2-1\right)}^{\frac{5}{2}}\left(3x^2\right)+\left(x^3+5\right)\left(\frac{5}{2}\right)\color{red}{\left(x^2-1\right)}^{\frac{3}{2}}\left(2x\right)\]
jim_thompson5910
  • jim_thompson5910
Factor it out to get \[\large \color{red}{\left(x^2-1\right)}^{\frac{5}{2}}\left(3x^2\right)+\left(x^3+5\right)\left(\frac{5}{2}\right)\color{red}{\left(x^2-1\right)}^{\frac{3}{2}}\left(2x\right)\] \[\large \color{red}{\left(x^2-1\right)}^{\frac{3}{2}}\left(\left(x^2-1\right)\left(3x^2\right)+\left(x^3+5\right)\left(\frac{5}{2}\right)\left(2x\right)\right)\] as you can see, it's not much of a simplification
jim_thompson5910
  • jim_thompson5910
well now that I think about it more, the 5/2 and 2x multiply to 5x so that's something
anonymous
  • anonymous
Question: How come the 3/2 exponent was kept but not the 5/2 ?
jim_thompson5910
  • jim_thompson5910
I factored out the portion with the 3/2 exponent
jim_thompson5910
  • jim_thompson5910
|dw:1444358829913:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1444358856055:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1444358869946:dw|
jim_thompson5910
  • jim_thompson5910
that's a similar example
anonymous
  • anonymous
Okay, I think I understand. Is that the only factoring that can be done?
jim_thompson5910
  • jim_thompson5910
yeah I think so, but again, it's not much of a simplification
jim_thompson5910
  • jim_thompson5910
I guess you could do this \[\large \color{black}{\left(x^2-1\right)}^{\frac{3}{2}}\left(\left(x^2-1\right)\left(3x^2\right)+\left(x^3+5\right)\left(\frac{5}{2}\right)\left(2x\right)\right)\] \[\large \color{black}{\left(x^2-1\right)}^{\frac{3}{2}}\left((x^2-1)(3x^2)+(x^3+5)(5x)\right)\] \[\large \color{black}{\left(x^2-1\right)}^{\frac{3}{2}}\left(3x^3-3x^2+5x^4+25x\right)\] \[\large \color{black}{\left(x^2-1\right)}^{\frac{3}{2}}\left(35x^4+x^3-3x^2+25x\right)\]
jim_thompson5910
  • jim_thompson5910
sorry typo, let me fix
jim_thompson5910
  • jim_thompson5910
\[\large \color{black}{\left(x^2-1\right)}^{\frac{3}{2}}\left(\left(x^2-1\right)\left(3x^2\right)+\left(x^3+5\right)\left(\frac{5}{2}\right)\left(2x\right)\right)\] \[\large \color{black}{\left(x^2-1\right)}^{\frac{3}{2}}\left((x^2-1)(3x^2)+(x^3+5)(5x)\right)\] \[\large \color{black}{\left(x^2-1\right)}^{\frac{3}{2}}\left(3x^4-3x^2+5x^4+25x\right)\] \[\large \color{black}{\left(x^2-1\right)}^{\frac{3}{2}}\left(8x^4-3x^2+25x\right)\]
anonymous
  • anonymous
Oh, that's perfect! Thank you yet again!
anonymous
  • anonymous
Now, I am off to learn about Extrema Derivatives. Wish me luck!

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