Question

for f(x)= [sin(x)]/x find f'(x) and f'(c) when c= pi/6
I am stuck on the second part!

Answer 1

Ok, so I found the first part:
f'(x)= [(cosx)x - (sinx)]/x^2

So for the second part, I have to put pi/6 in:
f'(pi/6)= [(cospi/6)pi/6 - sinpi/6]/(pi/6)^2
But I am not sure what to do. I could solve with my calculator but I think my teacher wants it in radian form. Can someone point me in the right direction?

Answer 2

Do you know the unit circle?

Answer 3

http://www.google.com/url?sa=i&source=images&cd=&cad=rja&docid=d6VR7E5XbmcHeM&tbnid=b6OyxY_TLZ5-KM:&ved=&url=http%3A%2F%2Fwww.sciencedigest.org%2Funit%2520circle.htm&ei=SRhwUreyGcbK2wXx0oDwDA&psig=AFQjCNHhNQ97n4oY97T2Gl_qGnOLlYfQTA&ust=1383164361456030

Answer 4

Not by heart, but I have a picture on my computer. I'll pull it up.

Answer 5

Use that to evaluate cos(pi/6) and sin(pi/6)

Answer 6

Would it be sqrt3/2?

Answer 7

That is for cos(pi/6)
and sin(pi/6)=1/2 right?

Answer 8

\[f'(\frac{\pi}{6})=\frac{\frac{\sqrt{3}}{2} \cdot \frac{\pi}{6}-\frac{1}{2}}{(\frac{\pi}{6})^2}\]
So this is what you have.
Most teachers do not like compound fractions.
Do you know how to clear the compound fractions?

Answer 9

Right, so sqrt3/2(pi/6) -1/2 all over pi/6^2 So what next?

Answer 10

look at the bottoms of the mini-fractions.
The highest number in them is the 36, but all the denominators of the mini-fractions go into 36 so multiply top and bottom of the big fraction by 36.

Answer 11

36? Sorry, but I'm a bit lost. Where did you get that from?

Answer 12

The 36 on bottom of the mini-fraction in the big fraction.