How do you find the derivative of $f(x)=3sec^7x$ ?

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Answer 1 · 2016-08-04T12:09:08

$=21sec^7(x)tan(x)$

$sec^7(x) = (sec(x))^7$

We are going to use the chain rule with

$u = sec(x) and y = 3u^7$

$(du)/(dx) = sec(x)tan(x) and (dy)/(du) = 21u^6$

$(dy)/(dx) = (dy)/(du)(du)/(dx)$

Multiplying and subbing back in for $u$ gives:

$(dy)/(dx) = 21(sec(x))^6*sec(x)tan(x)$

$=21sec^7(x)tan(x)$

Answer 2 · 2016-09-04T08:46:43

$21 sec^(7)(x) tan(x)$

We have: $f(x)=3 sec^(7)(x)$

This function can be differentiated using the "chain rule".

Let $u = sec(x) => u' = 3 sec(x) tan(x)$ and $v = 3 u^(7) => v' = 21 u^(6)$ :

$=> f'(x) = sec(x) tan(x) cdot 21 u^(6)$

$=> f'(x) = 21 sec(x) tan(x) u^(6)$

We can now replace $u$ with sec(x)#:

$=> f'(x) = 21 sec(x) tan(x) (sec(x))^(6)$

$=> f'(x) = 21 sec^(7)(x) tan(x)$

How do you find the derivative of f(x)=3sec^7xf(x)=3sec^7x?