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pottersheep

Having trouble with quotient rule - Calc help please. (x^2 + 1)^4 / x^4

  • one year ago
  • one year ago

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  1. pottersheep
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    I have to derive that

    • one year ago
  2. pottersheep
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    The answer is supposed to be [4(x^2-1)(x^2+1)]/x^5 which is not what I get

    • one year ago
  3. pottersheep
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    sorry the answer is [4(x^2-1)(x^2+1)^3]/x^5

    • one year ago
  4. ZeHanz
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    First write it in the equation editor to get a better look at it:\[\frac{ (x^2+1)^4 }{ x^4 }=\left( \frac{ x^2+1 }{ x } \right)^4\] Now you can also use the Chain Rule, because we have the 4th power as a second step. The derivative is:\[4\left( \frac{ x^2+1 }{ x } \right)^3 \cdot \frac{ x \cdot 2x - 1 \cdot(x^2+1) }{ x^2 }=4\left( \frac{ x^2+1 }{ x } \right)^3 \cdot \frac{ x^2- 1}{ x^2 }\]

    • one year ago
  5. pottersheep
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    Can I use chain rule where there is a fraction?

    • one year ago
  6. ZeHanz
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    Although we're done differentiating, we could still simplify:\[\frac{ 4(x^2+1)^3(x^2-1) }{ x^5 }\]

    • one year ago
  7. ZeHanz
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    @pottersheep: as long as there is a chain, you can (have to) use the Chain Rule! The chain here is: 1. u=(x^2+1)/x 2. y=u^4

    • one year ago
  8. pottersheep
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    Ah ok

    • one year ago
  9. pottersheep
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    Thank you for your explanation

    • one year ago
  10. ZeHanz
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    You could have done it without Chain Rule, if you leave the original function as it was:\[\left( \frac{ (x^2 + 1)^4 }{ x^4 }\right)'=\frac{ x^4 \cdot 4(x^2+1)^3 \cdot 2x-(x^2+1)^4 \cdot 4x^3 }{ x^8 }\] (still used Chain Rule - see factor 2x!) This now also has to be simplified (begin with dividing everything by x³):\[\frac{ 8x^2(x^2+1)^3-4(x^2+1)^4 }{ x^5 }=\]\[\frac{ 4(x^2+1)^3(2x^2-(x^2+1)) }{ x^5 }=\frac{ 4(x^2+1)^3(x^2-1) }{ x^5 }\]So we come to the same answer, eventually. You decide which method is easier!

    • one year ago
  11. pottersheep
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    OHhhhh that's what I did before! I see my mistake now! :)

    • one year ago
  12. pottersheep
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    Thanks again!

    • one year ago
  13. ZeHanz
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    yw!

    • one year ago
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