v.s
Differentiate.
f(θ) = sec θ/3 + sec θ



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Coolsector
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\[f(t) = \frac{ \sec(t) }{ 3 } + \sec(t)\]

Coolsector
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this is the question ?

Coolsector
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or it's 3+sec(t) in the denominator ?

v.s
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yeaaa

v.s
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in the denominator

Coolsector
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\[f(t)=\frac{ \sec(t) }{ 3 + \sec(t) }\]

Coolsector
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like this ?

v.s
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yes

micahwood50
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In this case, use quotient rule.

Coolsector
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grr mistake : all over again

Coolsector
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\[f'(t) = \frac{ \sec(t)' * (3+\sec(t))  (3+\sec(t))' * \sec(t) }{ (3+\sec(t))^2 }\]

Coolsector
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\[f'(t) = \frac{ \tan(t)\sec(t) * (3+\sec(t))  \tan(t)\sec(t) * \sec(t) }{ (3+\sec(t))^2 }\]

Coolsector
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\[f'(t) = \frac{ 3\tan(t)\sec(t) }{ (3+\sec(t))^2 }\]

Coolsector
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nothing much more to do ..

v.s
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so you expand and simplify?

Coolsector
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the final answer ? you can expand the (3+sec(t))^2 but it's not really important ..

Coolsector
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i used the quotient rule in order to find the derivative as you can see the procedure

v.s
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okaay thank you :)

Coolsector
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yw

v.s
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i got the aanswer wrong

v.s
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i'm doing this online quiz and wen i put in the answer it was wrong