## pottersheep Having trouble with quotient rule - Calc help please. (x^2 + 1)^4 / x^4 one year ago one year ago

1. pottersheep

I have to derive that

2. pottersheep

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

3. pottersheep

4. ZeHanz

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:$4\left( \frac{ x^2+1 }{ x } \right)^3 \cdot \frac{ x \cdot 2x - 1 \cdot(x^2+1) }{ x^2 }=4\left( \frac{ x^2+1 }{ x } \right)^3 \cdot \frac{ x^2- 1}{ x^2 }$

5. pottersheep

Can I use chain rule where there is a fraction?

6. ZeHanz

Although we're done differentiating, we could still simplify:$\frac{ 4(x^2+1)^3(x^2-1) }{ x^5 }$

7. ZeHanz

@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

8. pottersheep

Ah ok

9. pottersheep

10. ZeHanz

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!

11. pottersheep

OHhhhh that's what I did before! I see my mistake now! :)

12. pottersheep

Thanks again!

13. ZeHanz

yw!