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pottersheep
 one year ago
Best ResponseYou've already chosen the best response.0I have to derive that

pottersheep
 one year ago
Best ResponseYou've already chosen the best response.0The answer is supposed to be [4(x^21)(x^2+1)]/x^5 which is not what I get

pottersheep
 one year ago
Best ResponseYou've already chosen the best response.0sorry the answer is [4(x^21)(x^2+1)^3]/x^5

ZeHanz
 one year ago
Best ResponseYou've already chosen the best response.1First 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:\[4\left( \frac{ x^2+1 }{ x } \right)^3 \cdot \frac{ x \cdot 2x  1 \cdot(x^2+1) }{ x^2 }=4\left( \frac{ x^2+1 }{ x } \right)^3 \cdot \frac{ x^2 1}{ x^2 }\]

pottersheep
 one year ago
Best ResponseYou've already chosen the best response.0Can I use chain rule where there is a fraction?

ZeHanz
 one year ago
Best ResponseYou've already chosen the best response.1Although we're done differentiating, we could still simplify:\[\frac{ 4(x^2+1)^3(x^21) }{ x^5 }\]

ZeHanz
 one year ago
Best ResponseYou've already chosen the best response.1@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

pottersheep
 one year ago
Best ResponseYou've already chosen the best response.0Thank you for your explanation

ZeHanz
 one year ago
Best ResponseYou've already chosen the best response.1You 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):\[\frac{ 8x^2(x^2+1)^34(x^2+1)^4 }{ x^5 }=\]\[\frac{ 4(x^2+1)^3(2x^2(x^2+1)) }{ x^5 }=\frac{ 4(x^2+1)^3(x^21) }{ x^5 }\]So we come to the same answer, eventually. You decide which method is easier!

pottersheep
 one year ago
Best ResponseYou've already chosen the best response.0OHhhhh that's what I did before! I see my mistake now! :)
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