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

Answer 1

Paisley Johnson

#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)))#

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Answer 2

Evelyn Agard

To find the derivative of ( (x^2 + \frac{1}{x})^5 ), you would apply the chain rule. The derivative would be ( 5(x^2 + \frac{1}{x})^4 \times (2x - \frac{1}{x^2}) ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.