Question

Calculate the second derivative of the function.
f(θ) = sec^2(2θ)

Answer 1

first derivative is 4tan(2theta)sec^2(2theta)
right?

Answer 2

yes. it is right.

Answer 3

so u now have
\[ \large f'(\theta)=4\sec^2(2\theta)\tan(2\theta) \]
differentiate again.

Answer 4

this is what I got

Answer 5

it is wrong

Answer 6

i got
\[ \large f''(\theta)=16\sec^2(2\theta)\tan^2(2\theta)+8\sec^4(2\theta) \]

Answer 7

= 16 sec² (2x) tan² ( 2x) + 8 sec^4 (2x)

Answer 8

It'll be easier if you leave out the constant before taking derivative:
= 4 [ 4 sec² (2x) tan² ( 2x) + 2 sec^4 (2x) ]

Answer 9

Oh, yeah that's helpful.
I understand :) thank you

Answer 10

don't ever forget this rule:
\[ \large (\alpha f(x))'=\alpha f'(x) \]
usually, applying it easies things.

Answer 11

I don't understand that notation though, the other comment helped a lot

Answer 12

\=?
large=?