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

1 Answer
Jun 12, 2018

Answer:

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

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

Explanation:

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)#