Question

find the derivative:
f(x)= (x^2+1/x)^5

Answer 1

hey! i'm decent at doing these kinds of problems, itsj ust that ^5 outside confused me

Answer 2

my work so far got me this: 10(x^2+1)^4

Answer 3

the first one

Answer 4

\[((x^2+1)/x))^5\]

Answer 5

Avoid the quotient rule...$$\frac{x^2+1}x=\frac{x^2}x+\frac1x=x+x^{-1}$$

Answer 6

i applied the chain rule. i took the 5 and multiplied it by 2x ( the derivative inside)

Answer 7

That's not the derivative of the inside though!$$\frac{d}{dx}\left(x+x^{-1}\right)=1-x^{-2}$$

Answer 8

hold on my cats on my book

Answer 9

lol medal for a cat

Answer 10

did you take the x from the denom and change it to x^-1?

Answer 11

\(\frac1x=x^{-1}\) yep

Answer 12

ok so you have (x^-2+1)(X^-1)

Answer 13

Distribute the \(x^{-1}\) now.

Answer 14

i got x^-2 +x^-1

Answer 15

By the way it's \((x^2+1)(x^{-1})\). You should get \(x+x^{-1}\)

Answer 16

oh sorry! didn't realize i made that mistake

Answer 17

oh my gosh i was multipying the exponents

Answer 18

i was just supposed to subtract them :(

Answer 19

Just add the exponents, right. Anyways, if you follow what Callisto said, we now have...$$5(x+x^{-1})^4\times\frac{d}{dx}\left(x+x^{-1}\right)=5\left(x+x^{-1}\right)^4\times\left(1+(-1)x^{-2}\right)\\=5\left(x+\frac1x\right)^4\left(1-\frac1{x^2}\right)$$

Answer 20

so i got -5x +5(x+x^-1)^4

Answer 21

i made a mistake somewhere