anonymous
  • anonymous
Calculate the derivative of the function. Use Chain Rule g(z) = (z^2/7+z)^2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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baldymcgee6
  • baldymcgee6
Can you set it up?
anonymous
  • anonymous
im guessing its 2(z^2) x 2(7)
baldymcgee6
  • baldymcgee6
|dw:1351750568965:dw|Is this the function?

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anonymous
  • anonymous
@baldymcgee6 yes, and @sirm3d no the Z is squared
sirm3d
  • sirm3d
sorry i mistyped
anonymous
  • anonymous
yea i just have a hard time with this one. maybe because its a different format
sirm3d
  • sirm3d
the first part is differentiating the square
sirm3d
  • sirm3d
\[2\left( \frac{ z^{2} }{ 7+z } \right)\]
baldymcgee6
  • baldymcgee6
There is an easier method if you want..
baldymcgee6
  • baldymcgee6
|dw:1351750799134:dw|Just use the power rule and chain rule throughout, no need for quotient rule.
sirm3d
  • sirm3d
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
baldymcgee6
  • baldymcgee6
|dw:1351750960037:dw|
baldymcgee6
  • baldymcgee6
You will actually need the product rule because the co-efficient is a variable.
anonymous
  • anonymous
@sirm3d so the derivative would be 2x/7 ?and @baldymcgee6 so the previous equation you typed is the product rule? or chain?
sirm3d
  • sirm3d
nope. the quotient rule on \[\frac{ z^{2} }{ 7+z }\] will produce \[\frac{ \left( 7+z \right)\left( 2z \right)-\left( z^{2}\left( 1 \right) \right) }{ \left( 7+z \right)^{2} }\]
baldymcgee6
  • baldymcgee6
|dw:1351751195729:dw|
anonymous
  • anonymous
@baldymcgee6 quick question. where did u get (-1) in the very first step in the equation
baldymcgee6
  • baldymcgee6
|dw:1351751807412:dw|
anonymous
  • anonymous
yes
baldymcgee6
  • baldymcgee6
|dw:1351751909040:dw|
anonymous
  • anonymous
also the second step the last part of the equation is cut off -2(7+z....
baldymcgee6
  • baldymcgee6
sorry, my drawings keep getting cut off for some reason. -2(7+z)^-3 Just using power rule there.
anonymous
  • anonymous
awesome, thanks a lot
baldymcgee6
  • baldymcgee6
Welcome =D

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