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chrislb22 Group Title

Calculate the derivative of the function. Use Chain Rule g(z) = (z^2/7+z)^2

  • 2 years ago
  • 2 years ago

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  1. baldymcgee6 Group Title
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    Can you set it up?

    • 2 years ago
  2. chrislb22 Group Title
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    im guessing its 2(z^2) x 2(7)

    • 2 years ago
  3. baldymcgee6 Group Title
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    |dw:1351750568965:dw|Is this the function?

    • 2 years ago
  4. chrislb22 Group Title
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    @baldymcgee6 yes, and @sirm3d no the Z is squared

    • 2 years ago
  5. sirm3d Group Title
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    sorry i mistyped

    • 2 years ago
  6. chrislb22 Group Title
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    yea i just have a hard time with this one. maybe because its a different format

    • 2 years ago
  7. sirm3d Group Title
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    the first part is differentiating the square

    • 2 years ago
  8. sirm3d Group Title
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    \[2\left( \frac{ z^{2} }{ 7+z } \right)\]

    • 2 years ago
  9. baldymcgee6 Group Title
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    There is an easier method if you want..

    • 2 years ago
  10. baldymcgee6 Group Title
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    |dw:1351750799134:dw|Just use the power rule and chain rule throughout, no need for quotient rule.

    • 2 years ago
  11. sirm3d Group Title
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    the second part is where the chain rule is used. Get the derivative of \[\frac{ z^{2} }{ 7+z }\] by the quotient rule for differentiation

    • 2 years ago
  12. baldymcgee6 Group Title
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    |dw:1351750960037:dw|

    • 2 years ago
  13. baldymcgee6 Group Title
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    You will actually need the product rule because the co-efficient is a variable.

    • 2 years ago
  14. chrislb22 Group Title
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    @sirm3d so the derivative would be 2x/7 ?and @baldymcgee6 so the previous equation you typed is the product rule? or chain?

    • 2 years ago
  15. sirm3d Group Title
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    nope. the quotient rule on \[\frac{ z^{2} }{ 7+z }\] will produce \[\frac{ \left( 7+z \right)\left( 2z \right)-\left( z^{2}\left( 1 \right) \right) }{ \left( 7+z \right)^{2} }\]

    • 2 years ago
  16. baldymcgee6 Group Title
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    |dw:1351751195729:dw|

    • 2 years ago
  17. chrislb22 Group Title
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    @baldymcgee6 quick question. where did u get (-1) in the very first step in the equation

    • 2 years ago
  18. baldymcgee6 Group Title
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    |dw:1351751807412:dw|

    • 2 years ago
  19. chrislb22 Group Title
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    yes

    • 2 years ago
  20. baldymcgee6 Group Title
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    |dw:1351751909040:dw|

    • 2 years ago
  21. chrislb22 Group Title
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    also the second step the last part of the equation is cut off -2(7+z....

    • 2 years ago
  22. baldymcgee6 Group Title
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    sorry, my drawings keep getting cut off for some reason. -2(7+z)^-3 Just using power rule there.

    • 2 years ago
  23. chrislb22 Group Title
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    awesome, thanks a lot

    • 2 years ago
  24. baldymcgee6 Group Title
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    Welcome =D

    • 2 years ago
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