What is d/(d theta) (sec^2 4theta)ddθ(sec24θ) ?

2 Answers
Jul 23, 2016

d/(d theta) (sec^2 4theta) = 8sec^2 4theta tan 4thetaddθ(sec24θ)=8sec24θtan4θ

Explanation:

d/(d theta) (sec^2 4theta) = 2sec 4theta * d/(d theta) (sec 4 theta)ddθ(sec24θ)=2sec4θddθ(sec4θ) (Power rule and Chain rule)

= 2sec 4theta * tan 4theta sec 4theta * 4=2sec4θtan4θsec4θ4 (Standard differential and Chain rule)

= 8 sec^2 4theta tan 4theta=8sec24θtan4θ

Apr 4, 2018

8sec^2\4thetatan4theta8sec24θtan4θ

Explanation:

Given: d/(d\theta)(sec^2\4theta)ddθ(sec24θ)

We use the chain rule, which states that:

dy/dx=dy/(du)*(du)/dxdydx=dydududx

In this case, it's:

dy/(d\theta)=dy/(du)*(du)/(d\theta)dydθ=dydududθ

And so, let u=4thetau=4θ, 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