+9673 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?
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
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.
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.
+118673 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?
+118673 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. :(
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
\(\displaystyle \frac{d}{dx}\sin \left( \frac{1}{x}\right)=\lim_{h \rightarrow 0}\left(\frac{1}{h}\right)\left[\sin\left(\frac{1}{x+h}\right)-\sin\left(\frac{1}{x}\right)\right]\),
\(\displaystyle =\lim_{h \rightarrow 0}\left(\frac{1}{h}\right)\left[2\cos\left( \frac {2x+h}{2x(x + h)}\right)\sin\left(\frac{-h}{2x(x + h)}\right)\right]\),
(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
