if g(x)=x^2f'(x)

- amy0799

if g(x)=x^2f'(x)

- katieb

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- amy0799

where \[f(x)=\frac{ x-cosx }{ x }\] find the average rate of change of g on [pi, 2pi]

- SolomonZelman

differentiate the f(x).

- SolomonZelman

\(\large\color{black}{ \displaystyle f(x) =\frac{x-\cos x}{x} }\)
\(\large\color{black}{ \displaystyle f(x) =1-\frac{\cos x}{x} }\)
then derivative of 1 is 0, and derivative of the second peace will go by quotient rule (or you can use product or logarithmic)

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## More answers

- SolomonZelman

\(\large\color{black}{ \displaystyle f'(x) =0-\frac{(\cos x)'x-\cos x(x)'}{x^2} }\)

- SolomonZelman

continue please.

- amy0799

why did you take the 2nd derivative?

- SolomonZelman

I didn't take the second derivative.

- amy0799

oh wait I misread ur sentence, sorry

- SolomonZelman

My function is:
\(\large\color{black}{ \displaystyle f(x) =\frac{x-\cos x}{x} }\)
then I simplify it algebraically, to get:
\(\large\color{black}{ \displaystyle f(x) =1-\frac{\cos x}{x} }\)
And now, We have to differentiate it (once).
\(\large\color{black}{ \displaystyle f'(x) =0-\frac{(\cos x)'x-\cos x(x)'}{x^2} }\)

- SolomonZelman

if it is not making sense, then you are always welcome to express your concerns regarding the problem.

- amy0799

so then
\[g(x)=x ^{2}(\frac{ xsinx+cosx}{ x ^{2} })\]

- amy0799

- SolomonZelman

yes, very good, and you can tell that x² will can cancel

- SolomonZelman

Then the remaining part is an algebraic task of fining the slope (of the secant)
between \((\) 2π,f(2π) \()\) and \((\) π,f(π) \()\)

- amy0799

I put 2pi and pi into xsinx+cosx?

- amy0799

- SolomonZelman

Yes, that is what you do:)

- amy0799

(2pi,1) (pi,-1)
is that correct?

- SolomonZelman

Yes

- SolomonZelman

then the slope formula...

- amy0799

pi/2 is the answer?

- SolomonZelman

|dw:1442437619280:dw|

- SolomonZelman

Not the other way around

- amy0799

oh oops so 2/pi?

- SolomonZelman

Yup

- amy0799

do u mind helping me with another problem?

- SolomonZelman

Yes, if I will be able to.

- amy0799

find \[\lim_{h \rightarrow 0} \frac{ \csc(\frac{ \pi }{ 6 }+h)-\csc\frac{ \pi }{ 6 }}{ h } \]
@SolomonZelman

- SolomonZelman

Oh boy, first principles of differentiation.... (you will get to the point, when this will be only useful to perhaps prove derivtive of e^x, but I guess right now we can't apply the easy properties right... will go with that limit definition... i have to refresh my page. sorry)

- amy0799

when I did this I got 2 and I'm not sure if that's correct

- SolomonZelman

Something was going on with my PC, it is kind of off.

- SolomonZelman

Ok, I think I am back.

- SolomonZelman

With \(f(x)=\csc(x)\)
\(\large\color{slate}{\displaystyle f'(x)=\lim_{h \rightarrow ~0}\frac{\csc(x+h)-\csc(x)}{h}}\)
\(\large\color{slate}{\displaystyle f'(\pi/6)=\lim_{h \rightarrow ~0}\frac{\csc(\pi/6+h)-\csc(\pi/6)}{h}}\)

- SolomonZelman

You don't get 2 for that. I really don't think so.

- SolomonZelman

I get -2√3

- amy0799

how?

- SolomonZelman

I applied the chain rule:
\(\displaystyle \frac{d}{dx} (\csc x)=\frac{d}{dx} \left((\sin x)^{-1}\right)=(-1)\sin^{-2}x \times \color{blue}{\cos x} \\=-\csc^2x \cos x=-\cot x\csc x\)

- SolomonZelman

Then I plugged π/6 instead of x.

- SolomonZelman

This is just for check...

- SolomonZelman

But, I guess you have to use that lim h-> 0 rule, right?

- amy0799

where did you get the cosx from?

- SolomonZelman

have you studied the "Chain Rule" ?

- SolomonZelman

If not, then don't worry about my application of the chain rule...

- amy0799

yes

- SolomonZelman

ok, so you want to do the limit?

- amy0799

oh wait I see how you got it.

- SolomonZelman

\(\large\color{slate}{\displaystyle \lim_{h \rightarrow ~0}\frac{\csc(\frac{\pi}{6}+h)-\csc(\frac{\pi}{6})}{h}}\)

- SolomonZelman

Still want to do the limit?

- amy0799

sure. I don't understand how you got \[-\frac{ 2 }{ \sqrt{3} }\]

- SolomonZelman

I didn't.

- SolomonZelman

I said tha I got -2√3

- SolomonZelman

We have to do the limit, don't we?

- amy0799

I believe so

- SolomonZelman

\(\large\color{slate}{\displaystyle \lim_{h \rightarrow ~0}\frac{\csc(x+h)-\csc(x)}{h}}\)
\(\large\color{slate}{\displaystyle \lim_{h \rightarrow ~0}\frac{\dfrac{1}{\sin(x+h)}-\dfrac{1}{\sin(x)}}{h}}\)
\(\large\color{slate}{\displaystyle \lim_{h \rightarrow ~0}\frac{\dfrac{\sin(x)-\sin(x+h) }{\sin(x+h)\sin(x)}}{h}}\)
\(\large\color{slate}{\displaystyle \lim_{h \rightarrow ~0}\frac{\dfrac{\sin(x)-\sin(x)\cos(h) -\sin(h)\cos(x)}{[\sin(x)\cos(h) +\sin(h)\cos(x)]\sin(x)}}{h}}\)

- SolomonZelman

this is too much for me... going through all of this limit.

- amy0799

I'm sorta getting lost on what ur doing. does finding the limit get u \[-2\sqrt{3}?\]

- SolomonZelman

|dw:1442439909186:dw|

- SolomonZelman

If, finding the derivative through the chain rule, and plugging π/6 gives -2√3, then the limit must give the same.

- SolomonZelman

I drew two identities that you will need to get rid of h in the denominator....
It is just not right tho, if I will type all of this for you, according to the site's policy-;9

- amy0799

oh then I guess we don't have to find the limit. can you explain how you −2√3 from the chain rule?

- amy0799

how did you get −cotxcscx from csc^2xcosx?

- SolomonZelman

Do you get the essence of the chain rule?
(before we go on to our problem)

- amy0799

yes

- SolomonZelman

Ok, so we are differentiating \(\csc x\) (with respect to x)
we know that: \(\csc x = 1/\sin x = \left( \sin x \right)^{-1} \) right?

- amy0799

right

- SolomonZelman

So, we are going to differentiate *THAT* now.

- SolomonZelman

\(\large\color{black}{ \displaystyle \frac{d }{dx} \left[\left( \sin x \right)^{-1}\right] }\)

- SolomonZelman

Just as,
\(\large\color{black}{ \displaystyle \frac{d }{dx} \left[ x ^{-1}\right]=(-1)\cdot x^{-1-1}=-x^{-2} }\)
So is,
\(\large\color{black}{ \displaystyle \frac{d }{dx} \left[\left( \sin x \right)^{-1}\right] =(-1)\cdot \left( \sin x \right)^{-1-1}=-\left( \sin x \right)^{-2}}\)
However, there is a *BUT* to that.

- SolomonZelman

We must multiply times the derivative of the inner function, times the derivative of sin x.

- SolomonZelman

\(\large\color{black}{ \displaystyle \frac{d }{dx} \left[\left( \sin x \right)^{-1}\right] =(-1)\cdot \left( \sin x \right)^{-1-1} \color{blue}{\times \frac{d }{dx}\left[\sin x\right]}}\)

- SolomonZelman

this in blue, is the CHAIN RULE, that we must apply

- SolomonZelman

\(\large\color{black}{ \displaystyle \frac{d }{dx} \left[\left( \sin x \right)^{-1}\right] =(-1)\cdot \left( \sin x \right)^{-1-1} \color{blue}{\times \frac{d }{dx}\left[\sin x\right]}}\)
\(\large\color{black}{ \displaystyle \frac{d }{dx} \left[\left( \sin x \right)^{-1}\right] =- \left( \sin x \right)^{-2} \color{blue}{\times \cos x}}\)

- SolomonZelman

so far so good?

- amy0799

yup

- SolomonZelman

then, you just simplify it:
\(\large\color{black}{ \displaystyle \frac{d }{dx} \left[\left( \sin x \right)^{-1}\right] =- \left( \sin x \right)^{-2} \color{blue}{\times \cos x}}\)
\(\large\color{black}{ \displaystyle \frac{d }{dx} \left[\left( \sin x \right)^{-1}\right] =- \csc^2 x \color{blue}{\times \cos x}}\)
and thus we know that:
\(\large\color{black}{ \displaystyle \frac{d }{dx} \left[\csc x\right] =- \csc x \cot x}\)

- SolomonZelman

then, you just plug in π/6, which should not be a problem.

- amy0799

I don't understand how you got −csc^2x×cosx

- SolomonZelman

well, sin\(^{-2}\)x is same as 1/sin²x, which is csc²x.
And the one in front came from the power rule from the very beginning

- SolomonZelman

And cos(x), as I explained, is the chain rule.

- amy0799

oh ok. so I have 2 more problems I need help with, can u help me? if not that's ok.

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