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Differentiate. f(θ) = sec θ/3 + sec θ

Mathematics
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\[f(t) = \frac{ \sec(t) }{ 3 } + \sec(t)\]
this is the question ?
or it's 3+sec(t) in the denominator ?

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Other answers:

yeaaa
in the denominator
\[f(t)=\frac{ \sec(t) }{ 3 + \sec(t) }\]
like this ?
yes
In this case, use quotient rule.
grr mistake :| all over again
\[f'(t) = \frac{ \sec(t)' * (3+\sec(t)) - (3+\sec(t))' * \sec(t) }{ (3+\sec(t))^2 }\]
\[f'(t) = \frac{ \tan(t)\sec(t) * (3+\sec(t)) - \tan(t)\sec(t) * \sec(t) }{ (3+\sec(t))^2 }\]
\[f'(t) = \frac{ 3\tan(t)\sec(t) }{ (3+\sec(t))^2 }\]
nothing much more to do ..
so you expand and simplify?
the final answer ? you can expand the (3+sec(t))^2 but it's not really important ..
i used the quotient rule in order to find the derivative as you can see the procedure
okaay thank you :)
yw
i got the aanswer wrong
i'm doing this online quiz and wen i put in the answer it was wrong

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