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What is the derivative of #sqrt(t^5) + root(4)(t^9)#?

Answer 1

Scarlett Allison

#f'(t) = 5/2 t^(3/2) + 9/4 t^(5/4)#

or #" "f'(t) = t^(5/4)/4 (10 root(4)(t) + 9)#

Given: #f(t) = sqrt(t^5) + root(4)(t^9)#

Since #sqrt(x) = x^(1/2) " and " root(4)(x) = x^(1/4)#:

#f(t) = t^(5/2) + t^(9/4)#

#f'(t) = 5/2 t^(5/2 - 2/2) + 9/4 t^(9/4 - 4/4) = 5/2 t^(3/2) + 9/4 t^(5/4)#

#f'(t) = 10/4 t^(6/4) + 9/4 t^(5/4)#

#f'(t) = 1/4t^(5/4) (10t^(1/4) + 9)#

#f'(t) = t^(5/4)/4 (10 root(4)(t) + 9)#

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Answer 2

Evelyn Agard

The derivative of sqrt(t^5) + root(4)(t^9) with respect to t is (5/2)t^(3/2) + (9/4)t^(5/4).

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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.