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,

$=8sec^2\4thetatan4theta$

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