How do you differentiate $f (t) = \sqrt[3]{1 + tan t}$ $f(t)=root3(1+tant)$ ?

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Answer 1 · 2017-02-23T02:56:00

$f'(t) = sec^2t/(3(1+tant)^(2/3))$

$f(t) = root3(1+tant)$

$= (1+tant)^(1/3)$

$f'(t) = 1/3(1+tant)^(-2/3) * d/dt(1+tant)$ [Power rule and Chain rule]

$= 1/3(1+tant)^(-2/3) * (0+ sec^2t)$ [Standard differential]

$=sec^2t/(3(1+tant)^(2/3))$

How do you differentiate f(t)=root3(1+tant)f(t)=3√1+tantf(t)=root3(1+tant)?