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

Question

4821 views around the world

You can reuse this answer
Creative Commons License

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)#