amy0799
  • amy0799
if g(x)=x^2f'(x)
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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amy0799
  • amy0799
where \[f(x)=\frac{ x-cosx }{ x }\] find the average rate of change of g on [pi, 2pi]
SolomonZelman
  • SolomonZelman
differentiate the f(x).
SolomonZelman
  • 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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SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle f'(x) =0-\frac{(\cos x)'x-\cos x(x)'}{x^2} }\)
SolomonZelman
  • SolomonZelman
continue please.
amy0799
  • amy0799
why did you take the 2nd derivative?
SolomonZelman
  • SolomonZelman
I didn't take the second derivative.
amy0799
  • amy0799
oh wait I misread ur sentence, sorry
SolomonZelman
  • 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
  • SolomonZelman
if it is not making sense, then you are always welcome to express your concerns regarding the problem.
amy0799
  • amy0799
so then \[g(x)=x ^{2}(\frac{ xsinx+cosx}{ x ^{2} })\]
amy0799
  • amy0799
@SolomonZelman
SolomonZelman
  • SolomonZelman
yes, very good, and you can tell that x² will can cancel
SolomonZelman
  • 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
  • amy0799
I put 2pi and pi into xsinx+cosx?
amy0799
  • amy0799
@SolomonZelman
SolomonZelman
  • SolomonZelman
Yes, that is what you do:)
amy0799
  • amy0799
(2pi,1) (pi,-1) is that correct?
SolomonZelman
  • SolomonZelman
Yes
SolomonZelman
  • SolomonZelman
then the slope formula...
amy0799
  • amy0799
pi/2 is the answer?
SolomonZelman
  • SolomonZelman
|dw:1442437619280:dw|
SolomonZelman
  • SolomonZelman
Not the other way around
amy0799
  • amy0799
oh oops so 2/pi?
SolomonZelman
  • SolomonZelman
Yup
amy0799
  • amy0799
do u mind helping me with another problem?
SolomonZelman
  • SolomonZelman
Yes, if I will be able to.
amy0799
  • amy0799
find \[\lim_{h \rightarrow 0} \frac{ \csc(\frac{ \pi }{ 6 }+h)-\csc\frac{ \pi }{ 6 }}{ h } \] @SolomonZelman
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
  • amy0799
when I did this I got 2 and I'm not sure if that's correct
SolomonZelman
  • SolomonZelman
Something was going on with my PC, it is kind of off.
SolomonZelman
  • SolomonZelman
Ok, I think I am back.
SolomonZelman
  • 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
  • SolomonZelman
You don't get 2 for that. I really don't think so.
SolomonZelman
  • SolomonZelman
I get -2√3
amy0799
  • amy0799
how?
SolomonZelman
  • 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
  • SolomonZelman
Then I plugged π/6 instead of x.
SolomonZelman
  • SolomonZelman
This is just for check...
SolomonZelman
  • SolomonZelman
But, I guess you have to use that lim h-> 0 rule, right?
amy0799
  • amy0799
where did you get the cosx from?
SolomonZelman
  • SolomonZelman
have you studied the "Chain Rule" ?
SolomonZelman
  • SolomonZelman
If not, then don't worry about my application of the chain rule...
amy0799
  • amy0799
yes
SolomonZelman
  • SolomonZelman
ok, so you want to do the limit?
amy0799
  • amy0799
oh wait I see how you got it.
SolomonZelman
  • SolomonZelman
\(\large\color{slate}{\displaystyle \lim_{h \rightarrow ~0}\frac{\csc(\frac{\pi}{6}+h)-\csc(\frac{\pi}{6})}{h}}\)
SolomonZelman
  • SolomonZelman
Still want to do the limit?
amy0799
  • amy0799
sure. I don't understand how you got \[-\frac{ 2 }{ \sqrt{3} }\]
SolomonZelman
  • SolomonZelman
I didn't.
SolomonZelman
  • SolomonZelman
I said tha I got -2√3
SolomonZelman
  • SolomonZelman
We have to do the limit, don't we?
amy0799
  • amy0799
I believe so
SolomonZelman
  • 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
  • SolomonZelman
this is too much for me... going through all of this limit.
amy0799
  • amy0799
I'm sorta getting lost on what ur doing. does finding the limit get u \[-2\sqrt{3}?\]
SolomonZelman
  • SolomonZelman
|dw:1442439909186:dw|
SolomonZelman
  • SolomonZelman
If, finding the derivative through the chain rule, and plugging π/6 gives -2√3, then the limit must give the same.
SolomonZelman
  • 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
  • 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
  • amy0799
how did you get −cotxcscx from csc^2xcosx?
SolomonZelman
  • SolomonZelman
Do you get the essence of the chain rule? (before we go on to our problem)
amy0799
  • amy0799
yes
SolomonZelman
  • 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
  • amy0799
right
SolomonZelman
  • SolomonZelman
So, we are going to differentiate *THAT* now.
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \frac{d }{dx} \left[\left( \sin x \right)^{-1}\right] }\)
SolomonZelman
  • 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
  • SolomonZelman
We must multiply times the derivative of the inner function, times the derivative of sin x.
SolomonZelman
  • 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
  • SolomonZelman
this in blue, is the CHAIN RULE, that we must apply
SolomonZelman
  • 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
  • SolomonZelman
so far so good?
amy0799
  • amy0799
yup
SolomonZelman
  • 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
  • SolomonZelman
then, you just plug in π/6, which should not be a problem.
amy0799
  • amy0799
I don't understand how you got −csc^2x×cosx
SolomonZelman
  • 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
  • SolomonZelman
And cos(x), as I explained, is the chain rule.
amy0799
  • 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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