anonymous
  • anonymous
Compute the derivative dy/dx [(x^3 + 8x)(x^2 − x)] | x= 2
Mathematics
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anonymous
  • anonymous
Compute the derivative dy/dx [(x^3 + 8x)(x^2 − x)] | x= 2
Mathematics
jamiebookeater
  • jamiebookeater
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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angela210793
  • angela210793
can u go on urself?|dw:1351364346994:dw|
anonymous
  • anonymous
the last part of the equation is cut off
baldymcgee6
  • baldymcgee6
|dw:1351368595224:dw|

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baldymcgee6
  • baldymcgee6
So all you have left to do is to compute the derivatives of (x^3+8x) and (x^2-x).... Lets start with x^3+8x: |dw:1351368680013:dw|
anonymous
  • anonymous
so 3x^2+ 8
baldymcgee6
  • baldymcgee6
|dw:1351368877079:dw|
anonymous
  • anonymous
2x-1
baldymcgee6
  • baldymcgee6
|dw:1351369084559:dw|
baldymcgee6
  • baldymcgee6
You should be able to put it all together now
anonymous
  • anonymous
so (x^3+8x)(2x-1) + (3x^2+8)(x^3+8x)
baldymcgee6
  • baldymcgee6
not quite...
baldymcgee6
  • baldymcgee6
|dw:1351369459007:dw|
anonymous
  • anonymous
the last part of the equation is cut off (2x- ...
baldymcgee6
  • baldymcgee6
2x-1

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