Question

Find f'(x) and f'(c). MEDAL WILL BE REWARDED!!!
Function: f(x)=(sinx)/x
Value of c: pi/6

Answer 1

Do you know the quotient rule?

Answer 2

If I can get the equation in line then yes

Answer 3

in line?

Answer 4

In final form

Answer 5

I don't understand what you mean. This is a quotient because there is a division sign. I'm asking do you know a particular rule. The quotient rule is the one I'm asking about.
If you have (f/g)' then by quotient rule this is [f'*g-g'*f]/g^2

Answer 6

So would I get (sinx-cosx)/x^2

Answer 7

\[(\frac{f}{g})'=\frac{f' \cdot g- g' \cdot f}{g^2}\]
no...
Like f is the top and your top here is sin(x)
and g is the bottom and your bottom is x
You are going to go in that new quotient I provided for you and plug those in.
you also need to plug in for f' and g'.

Answer 8

okay, so sinx(x)-x^2(sinx)/x^2

Answer 9

\[(\frac{\sin(x)}{x})'=\frac{(\sin(x))' \cdot x -(x)' \cdot \sin(x)}{x^2} \]
like this....see i replaced my g's in that formula with x and I replaced my f's with sin(x)

Answer 10

Those little prime things tells us we need to find the derivative though .

Answer 11

oh okay give me one sec ill decipher this...

Answer 12

Im stuck, if I start from scrath I get sin(X)

Answer 13

Ok... Do you understand how I get replaced the f items with sin(x) and the g items with x?

Answer 14

Yes

Answer 15

Ok so the only thing to do is just find the derivative of sin(x) to replace (sin(x))' and just to find the derivative x to replace (x)' in:
\[(\frac{\sin(x)}{x})'=\frac{(\sin(x))' \cdot x -(x)' \sin(x)}{x^2} \]

Answer 16

what is the derivative of sin(x)?

Answer 17

cos(x)

Answer 18

so now we
\[(\frac{\sin(x)}{x})'=\frac{\cos(x) \cdot x - (x)' \sin(x)}{x^2} \]

Answer 19

what is the derivative of x?

Answer 20

1

Answer 21

so this means we have
\[(\frac{\sin(x)}{x})'=\frac{x \cos(x) - 1 \sin(x)}{x^2} =\frac{xcos(x)-\sin(x)}{x^2} \]
since 1sin(x)=sin(x)

Answer 22

now your last direction says to evaluate this at pi/6

Answer 23

Before we move on, are there any questions?

Answer 24

no I understand this now! Lets keep moving! :)

Answer 25

\[(\frac{\sin(x)}{x})'|_{x=\frac{\pi}{6}}=\frac{\frac{\pi}{6} \cos(\frac{\pi}{6})-\sin(\frac{\pi}{6})}{{(\frac{\pi}6})^2} \]

Answer 26

Do you know how to use the unit circle to evaluate those trig parts?

Answer 27

No, that's my biggest problem!

Answer 28

http://www.google.com/imgres?imgurl=http://ibmathematics.tripod.com/sitebuildercontent/sitebuilderpictures/unitcircle.gif&imgrefurl=http://ibmathematics.tripod.com/id3.html&h=1772&w=2012&sz=56&tbnid=FyuDfKO6GiIszM:&tbnh=91&tbnw=103&zoom=1&usg=__BvBHLhljLWlrimq4M1UNggI2Mfg=&docid=octEdD9F24EAeM&sa=X&ei=sil9UomBOcn6kQf09YDYCg&ved=0CD0Q9QEwBA

Click that link and look at it with me.

Answer 29

Now you are looking on that circle for when pi/6 occurs.

Answer 30

Do you see pi/6?

Answer 31

Okay so what exactly am I looking for.....yes I do! at the first 30?

Answer 32

Do you see the ordered pair when we have pi/6 radians?

Answer 33

yes, is that my final answer?

Answer 34

That first value in there or the first coordinate is your cos(pi/6) value.
That second value in there or the second coordinate is your sin(pi/6) value.

Answer 35

so plug them in to the equation?

Answer 36

first coordinate meaning x-coordinate
second coordinate meaning y-coordinate

Answer 37

yep

Answer 38

tell me what you have when you plug them in

Answer 39

okay hold on with me....give me one sec

Answer 40

wait what will the equation look like?

Answer 41

Tell me what cos(pi/6) is looking on that unit circle?

Answer 42

radical3/2, 1/2

Answer 43

and 1/2 is our sin(pi/6) value?

Answer 44

both of those aren't cos(pi/6)

Answer 45

ok

Answer 46

so cos(pi/6) is sqrt(3)/2
and
sin(pi/6) is 1/2
now I'm going to go to my formula and replace those trig parts with these lovely fractions...

Answer 47

so would it be radical3/2(1/2) - (1/2)/radical3/2^2?

Answer 48

\[(\frac{\sin(x)}{x})'|_{x=\frac{\pi}{6}}=\frac{\frac{\pi}{6} \cos(\frac{\pi}{6})-\sin(\frac{\pi}{6})}{{(\frac{\pi}6})^2} \]
\[=\frac{\frac{\pi}{6} \cdot \frac{\sqrt{3}}{2}-\frac{1}{2}}{(\frac{\pi}{6})^2} \]

Answer 49

I replaced cos(pi/6) with sqrt(3)/2
and I replaced sin(pi/6) with 1/2
and did nothing else

Answer 50

okay, I see now, I had my equation written down wrong, so now I can just solve?

Answer 51

we don't call it solving
we call it evaluating/simplifying an expression for a certain value

Answer 52

but yes this can be simplified

Answer 53

okay so can I simplify?

Answer 54

\[\frac{ \frac{\sqrt{3} \pi}{12}-\frac{1}{2}}{\frac{\pi^2}{36}}\]
I would start by clearing the compound fractions...
Meaning let's get rid of the mini-fractions in our big fraction.

Answer 55

I get 3(pi radical3 -6)/6

Answer 56

over pi I mean

Answer 57

What can we multiply on top and bottom for our big (main) fraction so that we get rid of the the mini-fractions ?
Look at my last fraction I put up.

Answer 58

you can get rid of the pi's or the GCF

Answer 59

Look our mini-fractions denominators 12,2, and 36.
The lcm of 12, 2, and 36 sometimes we denote this as lcm(12,2,36)
is 36 correct?

Answer 60

yes

Answer 61

lcm meaning least common multiple

Answer 62

so multiply top and bottom of my main fraction by 36

Answer 63

\[\frac{36}{36} \cdot \frac{ \frac{\sqrt{3} \pi }{12} -\frac{1}{2}}{\frac{\pi^2}{36}} \]

Answer 64

well that cncles out the bottom 36 and leaves us with pi^2 so for the top, hold on...

Answer 65

so we can write this as:
\[\frac{ \frac{36 \cdot \sqrt{3} \pi }{12} - \cdot \frac{36 \cdot 1}{2}}{\frac{36 \cdot \pi^2}{36}}\]

Answer 66

I get 3radical3 - 1/2

Answer 67

unfortunately that isn't correct.

Answer 68

okay so can you work this out for me? I see what your doing but im probably making tiny mistakes along the way

Answer 69

\[\frac{\frac{\cancel{12} \cdot 3 \sqrt{3} \pi }{ \cancel{12}} -\frac{\cancel{2} \cdot 18 }{\cancel{2}}}{\frac{\cancel{36} \pi^2}{\cancel{36}}}\]

Answer 70

the oh reason we multiplied by 36 on top and bottom is so we could get rid of mini-fractions and we see here that it worked.

Answer 71

so we get 3radical3pi-18/pi^2

Answer 72

Ive gotta get going though, we have been at this for 45 minutes. Thank you for your help

Answer 73

Well if you type it in somewhere make sure you write
(3sqrt(3)pi-18)/pi^2

Like put parenthesis around the thing you want to be the numerator so people don't think that the numerator is just 18 for that one part of the expression.

Answer 74

\[\frac{3 \sqrt{3} \pi -18}{\pi^2} \]