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. :)

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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
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\[(x^2-1)^\frac{5}{2} (x^3+5)\]
what are the instructions?
Sorry, the directions are to find the derivative. (not sure why i didn't post that?)

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Other answers:

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

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