What is #d/(d theta) (sec^2 4theta)# ?

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Answer 1 · 2016-07-23T01:03:47

#d/(d theta) (sec^2 4theta) = 8sec^2 4theta tan 4theta#

#d/(d theta) (sec^2 4theta) = 2sec 4theta * d/(d theta) (sec 4 theta)# (Power rule and Chain rule)

#= 2sec 4theta * tan 4theta sec 4theta * 4# (Standard differential and Chain rule)

#= 8 sec^2 4theta tan 4theta#

Answer 2 · 2018-04-04T15:06:23

#8sec^2\4thetatan4theta#

Given: #d/(d\theta)(sec^2\4theta)#

We use the chain rule, which states that:

#dy/dx=dy/(du)*(du)/dx#

In this case, it's:

#dy/(d\theta)=dy/(du)*(du)/(d\theta)#

And so, let #u=4theta#, then #du=4 \ d\theta,(du)/(d\theta)=4#.

Also, #y=sec^2u#, then #dy/(du)=2sec^2utanu#.

Therefore, we get:

#dy/dx=2sec^2utanu*4#

#=8sec^2utanu#

Reversing back our substitution, we get,

#=8sec^2\4thetatan4theta#