Find the derivative of f(t)=(1+tant)^1/3?

In this case, since the function is the result of
composition of two functions, we'll compute the first deriative using the chain
rule.

The outside function is the power function, whose
superscript is 1/3 and the inside function is the expression 1 + tan
t.

The chain rule can be written as the following product
of terms:

f'(t) = derivative of outside function*inside
function*derivative of inside function

f'(t) = (1/3)*`(1 +
tan t)^(1/3 - 1)` * (1 + tan t)'

f'(t) = `(1 + tan
t)^(-2/3)` /3`cos^(2)` t

Since the power of numerator is
negative, we'll use the negative power rule and we'll re-write the
result:

f'(t) = 1/3`cos^(2)` t*`root(3)((1+tan
t)^(2))`

But 1/`cos^(2)` t = `sec^(2)`
t

The requested derivative of the given
function is

f'(t) = `sec^(2)`
t/3`root(3)((1+tan t)^(2))` .