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

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

ZeHanz
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0Can I use chain rule where there is a fraction?

ZeHanz
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0Thank you for your explanation

ZeHanz
 2 years 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
 2 years 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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