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1)How to differentiate \(x\sin(\dfrac{1}{x})\) using the first principles?

2)How to differentiate \(x^2\sin(\dfrac{1}{x})\) using the first principles?

3)How to differentiate \(x^3\sin(\dfrac{1}{x})\) using the first principles?

 

I am recently in a class for gifted students and the teacher told us to do this in the very first lesson of calculus......

I am not new to calculus though, it is just too complicated that I have no idea how to simplify the \(\dfrac{f(x+h) - f(x)}{h}\)

 

I tried to use \(\sin x - \sin y = 2\cos (\dfrac{x+y}{2})\sin (\dfrac{x-y}{2})\) on the numerator when I was doing the 1st question but that doesn't work because somehow you will find \(\dfrac{-2x}{h}\)something something + something that blocked your way.

 

Anyone please help me on this?

 Dec 22, 2016
edited by MaxWong  Dec 22, 2016
edited by MaxWong  Dec 22, 2016

Best Answer 

 #13
avatar+33603 
+10

Here's my take on 1):

 

 

For 2) I would start by writing  I2 = x*I1   and then for 3) by writing I3 = x*I2

 Dec 24, 2016
 #1
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+3

Hi Max:
I put this in Wolfram/Alpha and asked it to simplify it. This is what it came up with:

I don't know if this is helpful or not for your question.


 simplify step by step  f[(x + h)  - f(x)] / h
Possible derivation:
d/dx((f(h + x - f(x)))/h)
Factor out constants:
 = (d/dx(f(h + x - f(x))))/h
Using the chain rule, d/dx(f(-f(x) + h + x)) = ( df(u))/( du) 0, where u = -f(x) + h + x and ( d)/( du)(f(u)) = f'(u):
 = d/dx(h + x - f(x)) f'(h + x - f(x))/h
Differentiate the sum term by term and factor out constants:
 = d/dx(h) + d/dx(x) - d/dx(f(x)) (f'(h + x - f(x)))/h
The derivative of h is zero:
 = ((d/dx(x) - d/dx(f(x)) + 0) f'(h + x - f(x)))/h
Simplify the expression:
 = ((d/dx(x) - d/dx(f(x))) f'(h + x - f(x)))/h
The derivative of x is 1:
 = ((-(d/dx(f(x))) + 1) f'(h + x - f(x)))/h
The derivative of f(x) is f'(x):
Answer: | = ((1 - f'(x)) f'(h + x - f(x)))/h

 Dec 22, 2016
 #2
avatar+9466 
0

Your answer is not quite useful though...... But anyways thanks for attempting to help :)

MaxWong  Dec 22, 2016
 #3
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0

Max: "heureka", the German mathematican is excellent at this. I have seen him break this in detail a few months back. You could put the question to him directly, in English, on the German-equivalent of this site here: web2.0rechner.de

However, because of the upcoming Christmas holidays, he may be difficult to get. Good luck. 

 Dec 22, 2016
 #4
avatar+9466 
0

I know him...... But this is very urgent that I need to submit the homework tomorrow(about 9 hours later for you, and I have to sleep.)...... So I cannot wait for him to be online for a whole day.......

MaxWong  Dec 22, 2016
 #5
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+5

Hi Max,you realise that you are dealing with the product rule and chain rule here,but d/dx sin(1/x) 

=  { sin(1/x+h) - sin(1/x) }   /h.    Now use your formula sinA - sinB -2cos(A+B)/2 .sin(A-B)/2 to get

 

2cos{(x+1/2h)/x(x+1/2h) } .sin {(-1/2h)/ x(x+1/2h)

-----------------------------------------------------------------

 

               h               lim h tends to zero

 

divide numerator and denominator by 2    to get

 

cos{ (x+1/2h/x(x+1/2h) .sin{ (-1/2h)/x(x+1/2h)

-----------------------------------------------------------             since sin (x)/x tends to 1 in lim h tends to zero

 

         1/2 h

you are left with cos  { (x+ 1/2h)/x(x+1/2h)  which reduces to cos (x/x^2) as h tends to zero which gives cos(1/x).

You can go on to prove the product and chain rules to validate the rest of the derivative.Hope this helps. 

 Dec 22, 2016
 #6
avatar+118587 
+10

Hi guest 5

I do not see that you have answered the question.

The question asks you to differentiate   xsin(1/x)    not just   sin(1/x)

 

 

 

The answer that Wolfram alpha gives is

 

Looking better at what you hae written you have said that it is just a starting point, which is fair enough but 

I would like you to explain this statement of yours:

 

"  Now use your formula sinA - sinB -2cos(A+B)/2 .sin(A-B)/2   "

Where does this come from - there is not even an equal sign so how can it be a formula?

Melody  Dec 23, 2016
 #7
avatar+9466 
+5

Thanks Melody :)

MaxWong  Dec 23, 2016
 #8
avatar+118587 
+5

Hi Max,

I looked at your question but I am not good at these.  I have not seen Heureka in a while but hopefully he will drop in and give us an answer.  I love to learn how to do these from him.

OR Maybe guest 5 will come back and clarify what he was talking about because it didn't make any sense to me.  :(

Melody  Dec 23, 2016
 #9
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+10

Hi Melody

 

The trig formula for sinA - sinB is one of a group of four identities, the others being for sinA + sinB,

cosA + cosB and cosA - cosB. It helps if you can remember all four of them.

 

The identity for sinA + sinB is

sinA + sinB = 2sin{(A + B)/2}cos{(A - B)/2}

It's derived by adding the identities for sin(C + D) and sin(C - D) and then writing C + D = A and C - D = B.

I remember it as sin + sin = 2 sin semi sum cos semi diff. (if that makes sense !)

The others are

sin - sin = 2 cos semi sum sin semi diff

cos + cos = 2 cos semi sum cos semi diff

and

cos - cos = -2 sin semi sum sin semi diff.

(They are not difficult to remember if you look at the form of the identities for sin(A + B), sin(A - B) and so on.)

 

The mistake in #5 is that the negative sign in front of the first 2 should be an equals sign.

 

As to the calculus, I think that the original question is not well defined, are we supposed (to try) to diferentiate (1/x)sin(1/x) from first principles or are we allowed to derive the product and foaf rules and simply differentiate sin(x) from first principles, or do we derive just the product rule and differentiate sin(1/x) from first principles ?

(Can we assume the product and foaf rules ?)

 

If I have time, I'll post a derivation for sin(1/x) after lunch (UK).

 

Tiggsy

 Dec 23, 2016
 #10
avatar+118587 
+5

Thanks very much Tiggsy.

I shall look at what you have presented so far :))

Melody  Dec 23, 2016
 #11
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+10

\(\displaystyle \frac{d}{dx}\sin \left( \frac{1}{x}\right)=\lim_{h \rightarrow 0}\left(\frac{1}{h}\right)\left[\sin\left(\frac{1}{x+h}\right)-\sin\left(\frac{1}{x}\right)\right]\),

 

\(\displaystyle =\lim_{h \rightarrow 0}\left(\frac{1}{h}\right)\left[2\cos\left( \frac {2x+h}{2x(x + h)}\right)\sin\left(\frac{-h}{2x(x + h)}\right)\right]\),

 

(using the identity for sine minus sine).

 

The cosine term will tend to cos(1/x) as h tends to zero, so ignoring that for the moment, the rest of it can be written

 

\(\displaystyle - \lim_{h \rightarrow 0} \frac{1}{x(x+h)}\sin\left(\frac{h/2}{x(x+h)}\right)/\left(\frac{h/2}{x(x+h)}\right)\)

 

the x(x + h) being introduced so that we have a sin(theta)/theta ratio. (It cancels out).

 

and that will equal \(-1/x^{2}\) since the sin(theta)/theta part of it tends to 1 as h tends to zero.

 

Put that with the earlier cosine term and we have the final result \(-\frac{1}{x^{2}}\cos(\frac{1}{x})\).

 

Still not sure what to do about (1/x)sin(1/x) though.

 

Tiggsy

 Dec 23, 2016
 #12
avatar+118587 
+5

Thanks Tiggsy, 

This will give me lots to think about :))

 Dec 23, 2016
 #13
avatar+33603 
+10
Best Answer

Here's my take on 1):

 

 

For 2) I would start by writing  I2 = x*I1   and then for 3) by writing I3 = x*I2

Alan Dec 24, 2016
 #14
avatar+118587 
+10

Thanks Alan,

 

More brain food for me  :)))

Melody  Dec 24, 2016

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