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))` .