How do you find the derivative of #(x^2 + 1/x)^5#?

Question

1357 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-05T17:58:55

#5 (x^(2) + (1) / (x))^(4) (2 x - (1) / (x^(2)))#

We have: #(x^(2) + (1) / (x))^(5)#

This expression can be differentiated using the "chain rule".

Let #u = x^(2) + (1) / (x) => u' = 2 x - x^(- 2)# and #v = u^(5) => v' = 5 u^(4)#:

#=> (d) / (dx) ((x^(2) + (1) / (x))^(5)) = (2 x - x^(- 2)) cdot 5 u^(4)#

#=> (d) / (dx) ((x^(2) + (1) / (x))^(5)) = 5 u^(4) (2 x - (1) / (x^(2)))#

We can now replace #u# with #x^(2) + (1) / (x)#:

#=> (d) / (dx) ((x^(2) + (1) / (x))^(5)) = 5 (x^(2) + (1) / (x))^(4) (2 x - (1) / (x^(2)))#