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

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Answer 1 · 2018-06-12T03:17:37

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

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

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

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

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

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

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

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

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

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