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ShadowysBest ResponseYou've already chosen the best response.1
Have you differentiated F(x) and then sub'ed \(\frac{\pi}{6}\) into it?
 one year ago

waterineyesBest ResponseYou've already chosen the best response.1
Do you know the derivative of tan(x) ??
 one year ago

reeny1996Best ResponseYou've already chosen the best response.0
i kno the differvative is secxtanx but i dont know where to go frm there :/
 one year ago

waterineyesBest ResponseYou've already chosen the best response.1
Yes, than just replace x by pi/6..
 one year ago

reeny1996Best ResponseYou've already chosen the best response.0
(2\[\sqrt{3}\]/3 and \[\sqrt{3}\]/3
 one year ago

waterineyesBest ResponseYou've already chosen the best response.1
\[F'(\frac{\pi}{6}) = \sec(\frac{\pi}{6}) \cdot \tan(\frac{\pi}{6})\]
 one year ago

waterineyesBest ResponseYou've already chosen the best response.1
What is the value of : \[\sec (\frac{\pi}{6})\]
 one year ago

waterineyesBest ResponseYou've already chosen the best response.1
Yes and what is for tan(pi/6) = ??
 one year ago

waterineyesBest ResponseYou've already chosen the best response.1
Wait, derivative of tan(x) isn't it sec^2(x) ??
 one year ago

waterineyesBest ResponseYou've already chosen the best response.1
\[\frac{d}{dx} \tan(x) = \sec^2(x)\]
 one year ago

reeny1996Best ResponseYou've already chosen the best response.0
thts y im getting the wrong answer
 one year ago

waterineyesBest ResponseYou've already chosen the best response.1
You were doing the derivative of sec(x)..
 one year ago

waterineyesBest ResponseYou've already chosen the best response.1
\[\frac{d}{dx}\sec(x) = \sec(x) \cdot \tan(x)\] It is okay and sorry from my side too, I also got confused.. And you are welcome dear..
 one year ago

reeny1996Best ResponseYou've already chosen the best response.0
ok idk how to combine the 2 radicals
 one year ago

waterineyesBest ResponseYou've already chosen the best response.1
\[F'(\frac{\pi}{6}) = \sec^2(\frac{\pi}{6})\]
 one year ago

reeny1996Best ResponseYou've already chosen the best response.0
dw:1352970940060:dw
 one year ago

waterineyesBest ResponseYou've already chosen the best response.1
\[\implies \frac{2 \sqrt{3}}{3} \times \frac{2\sqrt{3}}{3} \implies ??\]
 one year ago

waterineyesBest ResponseYou've already chosen the best response.1
This is simple multiplication and division now...
 one year ago

reeny1996Best ResponseYou've already chosen the best response.0
i know but im not getting the right answer
 one year ago

waterineyesBest ResponseYou've already chosen the best response.1
What are you getting tell me first...
 one year ago

waterineyesBest ResponseYou've already chosen the best response.1
Its not 4... You are missing something.
 one year ago

waterineyesBest ResponseYou've already chosen the best response.1
\[\sqrt{3} \times \sqrt{3} = ??\]
 one year ago

reeny1996Best ResponseYou've already chosen the best response.0
2 times 2=4 root3 times root 3= 3 4 times 3 =12 12 divide3 =4
 one year ago

waterineyesBest ResponseYou've already chosen the best response.1
12 divided by 3 ??? Are you sure in denominator it is 3 ??
 one year ago

reeny1996Best ResponseYou've already chosen the best response.0
im not too good with my unit circle
 one year ago

waterineyesBest ResponseYou've already chosen the best response.1
Dear, what will you get when you will square 3 ??
 one year ago

waterineyesBest ResponseYou've already chosen the best response.1
Check the colored things here : \[\frac{2 \sqrt{3}}{\color{red}{3}} \times \frac{2 \sqrt{3}}{\color{red}{3}} = ??\]
 one year ago

waterineyesBest ResponseYou've already chosen the best response.1
Yes it is 12 divided by 9 and not 3.. So what will you get now ??
 one year ago

reeny1996Best ResponseYou've already chosen the best response.0
i was thinking addition
 one year ago

reeny1996Best ResponseYou've already chosen the best response.0
my mistake...srry if i took up yhur time
 one year ago

waterineyesBest ResponseYou've already chosen the best response.1
It is okay but be careful from next time... These are called "Silly Mistakes..." Ok, now concentrate more on new problems...
 one year ago
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