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

waterineyes
 2 years ago
Best ResponseYou've already chosen the best response.1Do you know the derivative of tan(x) ??

reeny1996
 2 years ago
Best ResponseYou've already chosen the best response.0i kno the differvative is secxtanx but i dont know where to go frm there :/

waterineyes
 2 years ago
Best ResponseYou've already chosen the best response.1Yes, than just replace x by pi/6..

reeny1996
 2 years ago
Best ResponseYou've already chosen the best response.0(2\[\sqrt{3}\]/3 and \[\sqrt{3}\]/3

waterineyes
 2 years ago
Best ResponseYou've already chosen the best response.1\[F'(\frac{\pi}{6}) = \sec(\frac{\pi}{6}) \cdot \tan(\frac{\pi}{6})\]

waterineyes
 2 years ago
Best ResponseYou've already chosen the best response.1What is the value of : \[\sec (\frac{\pi}{6})\]

waterineyes
 2 years ago
Best ResponseYou've already chosen the best response.1Yes and what is for tan(pi/6) = ??

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

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

reeny1996
 2 years ago
Best ResponseYou've already chosen the best response.0thts y im getting the wrong answer

waterineyes
 2 years ago
Best ResponseYou've already chosen the best response.1You were doing the derivative of sec(x)..

waterineyes
 2 years ago
Best 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..

reeny1996
 2 years ago
Best ResponseYou've already chosen the best response.0ok idk how to combine the 2 radicals

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

reeny1996
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1352970940060:dw

waterineyes
 2 years ago
Best ResponseYou've already chosen the best response.1\[\implies \frac{2 \sqrt{3}}{3} \times \frac{2\sqrt{3}}{3} \implies ??\]

waterineyes
 2 years ago
Best ResponseYou've already chosen the best response.1This is simple multiplication and division now...

reeny1996
 2 years ago
Best ResponseYou've already chosen the best response.0i know but im not getting the right answer

waterineyes
 2 years ago
Best ResponseYou've already chosen the best response.1What are you getting tell me first...

waterineyes
 2 years ago
Best ResponseYou've already chosen the best response.1Its not 4... You are missing something.

waterineyes
 2 years ago
Best ResponseYou've already chosen the best response.1\[\sqrt{3} \times \sqrt{3} = ??\]

reeny1996
 2 years ago
Best ResponseYou've already chosen the best response.02 times 2=4 root3 times root 3= 3 4 times 3 =12 12 divide3 =4

waterineyes
 2 years ago
Best ResponseYou've already chosen the best response.112 divided by 3 ??? Are you sure in denominator it is 3 ??

reeny1996
 2 years ago
Best ResponseYou've already chosen the best response.0im not too good with my unit circle

waterineyes
 2 years ago
Best ResponseYou've already chosen the best response.1Dear, what will you get when you will square 3 ??

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

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

reeny1996
 2 years ago
Best ResponseYou've already chosen the best response.0i was thinking addition

reeny1996
 2 years ago
Best ResponseYou've already chosen the best response.0my mistake...srry if i took up yhur time

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