if F(x)=tanx, then fprime(pi/6)=

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if F(x)=tanx, then fprime(pi/6)=

Calculus1
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Have you differentiated F(x) and then sub'ed \(\frac{\pi}{6}\) into it?
Do you know the derivative of tan(x) ??
i kno the differvative is secxtanx but i dont know where to go frm there :/

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Yes, than just replace x by pi/6..
(2\[\sqrt{3}\]/3 and \[\sqrt{3}\]/3
\[F'(\frac{\pi}{6}) = \sec(\frac{\pi}{6}) \cdot \tan(\frac{\pi}{6})\]
igot tht part
What is the value of : \[\sec (\frac{\pi}{6})\]
2root3/3
Yes and what is for tan(pi/6) = ??
root3/3
Wait, derivative of tan(x) isn't it sec^2(x) ??
its not ?
\[\frac{d}{dx} \tan(x) = \sec^2(x)\]
thts y im getting the wrong answer
thank you
You were doing the derivative of sec(x)..
yea
\[\frac{d}{dx}\sec(x) = \sec(x) \cdot \tan(x)\] It is okay and sorry from my side too, I also got confused.. And you are welcome dear..
ok idk how to combine the 2 radicals
Where ??
sec^2
\[F'(\frac{\pi}{6}) = \sec^2(\frac{\pi}{6})\]
yea
|dw:1352970940060:dw|
\[\implies \frac{2 \sqrt{3}}{3} \times \frac{2\sqrt{3}}{3} \implies ??\]
This is simple multiplication and division now...
i know but im not getting the right answer
4
What are you getting tell me first...
Its not 4... You are missing something.
\[\sqrt{3} \times \sqrt{3} = ??\]
2 times 2=4 root3 times root 3= 3 4 times 3 =12 12 divide3 =4
12 divided by 3 ??? Are you sure in denominator it is 3 ??
im not too good with my unit circle
Dear, what will you get when you will square 3 ??
9
Check the colored things here : \[\frac{2 \sqrt{3}}{\color{red}{3}} \times \frac{2 \sqrt{3}}{\color{red}{3}} = ??\]
9
Yes it is 12 divided by 9 and not 3.. So what will you get now ??
ohhhhhh
4/3
i was thinking addition
my mistake...srry if i took up yhur time
It is okay but be careful from next time... These are called "Silly Mistakes..." Ok, now concentrate more on new problems...
It is okay..
thanks a lot :)))

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