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amy0799
 one year ago
if g(x)=x^2f'(x)
amy0799
 one year ago
if g(x)=x^2f'(x)

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amy0799
 one year ago
Best ResponseYou've already chosen the best response.0where \[f(x)=\frac{ xcosx }{ x }\] find the average rate of change of g on [pi, 2pi]

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0differentiate the f(x).

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0\(\large\color{black}{ \displaystyle f(x) =\frac{x\cos x}{x} }\) \(\large\color{black}{ \displaystyle f(x) =1\frac{\cos x}{x} }\) then derivative of 1 is 0, and derivative of the second peace will go by quotient rule (or you can use product or logarithmic)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0\(\large\color{black}{ \displaystyle f'(x) =0\frac{(\cos x)'x\cos x(x)'}{x^2} }\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0continue please.

amy0799
 one year ago
Best ResponseYou've already chosen the best response.0why did you take the 2nd derivative?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0I didn't take the second derivative.

amy0799
 one year ago
Best ResponseYou've already chosen the best response.0oh wait I misread ur sentence, sorry

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0My 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
 one year ago
Best ResponseYou've already chosen the best response.0if it is not making sense, then you are always welcome to express your concerns regarding the problem.

amy0799
 one year ago
Best ResponseYou've already chosen the best response.0so then \[g(x)=x ^{2}(\frac{ xsinx+cosx}{ x ^{2} })\]

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0yes, very good, and you can tell that x² will can cancel

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0Then the remaining part is an algebraic task of fining the slope (of the secant) between \((\) 2π,f(2π) \()\) and \((\) π,f(π) \()\)

amy0799
 one year ago
Best ResponseYou've already chosen the best response.0I put 2pi and pi into xsinx+cosx?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0Yes, that is what you do:)

amy0799
 one year ago
Best ResponseYou've already chosen the best response.0(2pi,1) (pi,1) is that correct?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0then the slope formula...

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442437619280:dw

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0Not the other way around

amy0799
 one year ago
Best ResponseYou've already chosen the best response.0do u mind helping me with another problem?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0Yes, if I will be able to.

amy0799
 one year ago
Best ResponseYou've already chosen the best response.0find \[\lim_{h \rightarrow 0} \frac{ \csc(\frac{ \pi }{ 6 }+h)\csc\frac{ \pi }{ 6 }}{ h } \] @SolomonZelman

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0Oh 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
 one year ago
Best ResponseYou've already chosen the best response.0when I did this I got 2 and I'm not sure if that's correct

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0Something was going on with my PC, it is kind of off.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0Ok, I think I am back.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0With \(f(x)=\csc(x)\) \(\large\color{slate}{\displaystyle f'(x)=\lim_{h \rightarrow ~0}\frac{\csc(x+h)\csc(x)}{h}}\) \(\large\color{slate}{\displaystyle f'(\pi/6)=\lim_{h \rightarrow ~0}\frac{\csc(\pi/6+h)\csc(\pi/6)}{h}}\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0You don't get 2 for that. I really don't think so.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0Then I plugged π/6 instead of x.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0This is just for check...

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0But, I guess you have to use that lim h> 0 rule, right?

amy0799
 one year ago
Best ResponseYou've already chosen the best response.0where did you get the cosx from?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0have you studied the "Chain Rule" ?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0If not, then don't worry about my application of the chain rule...

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0ok, so you want to do the limit?

amy0799
 one year ago
Best ResponseYou've already chosen the best response.0oh wait I see how you got it.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0\(\large\color{slate}{\displaystyle \lim_{h \rightarrow ~0}\frac{\csc(\frac{\pi}{6}+h)\csc(\frac{\pi}{6})}{h}}\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0Still want to do the limit?

amy0799
 one year ago
Best ResponseYou've already chosen the best response.0sure. I don't understand how you got \[\frac{ 2 }{ \sqrt{3} }\]

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0I said tha I got 2√3

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0We have to do the limit, don't we?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0\(\large\color{slate}{\displaystyle \lim_{h \rightarrow ~0}\frac{\csc(x+h)\csc(x)}{h}}\) \(\large\color{slate}{\displaystyle \lim_{h \rightarrow ~0}\frac{\dfrac{1}{\sin(x+h)}\dfrac{1}{\sin(x)}}{h}}\) \(\large\color{slate}{\displaystyle \lim_{h \rightarrow ~0}\frac{\dfrac{\sin(x)\sin(x+h) }{\sin(x+h)\sin(x)}}{h}}\) \(\large\color{slate}{\displaystyle \lim_{h \rightarrow ~0}\frac{\dfrac{\sin(x)\sin(x)\cos(h) \sin(h)\cos(x)}{[\sin(x)\cos(h) +\sin(h)\cos(x)]\sin(x)}}{h}}\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0this is too much for me... going through all of this limit.

amy0799
 one year ago
Best ResponseYou've already chosen the best response.0I'm sorta getting lost on what ur doing. does finding the limit get u \[2\sqrt{3}?\]

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442439909186:dw

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0If, finding the derivative through the chain rule, and plugging π/6 gives 2√3, then the limit must give the same.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0oh then I guess we don't have to find the limit. can you explain how you −2√3 from the chain rule?

amy0799
 one year ago
Best ResponseYou've already chosen the best response.0how did you get −cotxcscx from csc^2xcosx?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0Do you get the essence of the chain rule? (before we go on to our problem)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0Ok, so we are differentiating \(\csc x\) (with respect to x) we know that: \(\csc x = 1/\sin x = \left( \sin x \right)^{1} \) right?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0So, we are going to differentiate *THAT* now.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0\(\large\color{black}{ \displaystyle \frac{d }{dx} \left[\left( \sin x \right)^{1}\right] }\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0Just as, \(\large\color{black}{ \displaystyle \frac{d }{dx} \left[ x ^{1}\right]=(1)\cdot x^{11}=x^{2} }\) So is, \(\large\color{black}{ \displaystyle \frac{d }{dx} \left[\left( \sin x \right)^{1}\right] =(1)\cdot \left( \sin x \right)^{11}=\left( \sin x \right)^{2}}\) However, there is a *BUT* to that.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0We must multiply times the derivative of the inner function, times the derivative of sin x.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0\(\large\color{black}{ \displaystyle \frac{d }{dx} \left[\left( \sin x \right)^{1}\right] =(1)\cdot \left( \sin x \right)^{11} \color{blue}{\times \frac{d }{dx}\left[\sin x\right]}}\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0this in blue, is the CHAIN RULE, that we must apply

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0\(\large\color{black}{ \displaystyle \frac{d }{dx} \left[\left( \sin x \right)^{1}\right] =(1)\cdot \left( \sin x \right)^{11} \color{blue}{\times \frac{d }{dx}\left[\sin x\right]}}\) \(\large\color{black}{ \displaystyle \frac{d }{dx} \left[\left( \sin x \right)^{1}\right] = \left( \sin x \right)^{2} \color{blue}{\times \cos x}}\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0then, 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
 one year ago
Best ResponseYou've already chosen the best response.0then, you just plug in π/6, which should not be a problem.

amy0799
 one year ago
Best ResponseYou've already chosen the best response.0I don't understand how you got −csc^2x×cosx

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0well, 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
 one year ago
Best ResponseYou've already chosen the best response.0And cos(x), as I explained, is the chain rule.

amy0799
 one year ago
Best ResponseYou've already chosen the best response.0oh ok. so I have 2 more problems I need help with, can u help me? if not that's ok.
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