Are you l33t at differentiating?

[tedz0r]

accid

[krtxmrtz]

ive been trying to get this one for ages but i cant...

differentiation by first principles on the function y=sin x

dy/dx = [2cos(x + h/2)sin(h/2)]/h

if you can actually do this one, itd be really good if u could show the trigonometric identities u used

thanks alot anyone who trys

If you're allowed to use the Taylor series for the sine then you're done, but I'm afraid this would be cheating...

[zaza]

Do you know how to differentiate by first principles?

df/dx = (f(x+dx) - f(x-dx))/2dx

You need only to know what sin(A+B) is and to recognise how sin(-A) relates to sin(A) and cos(-A) to cos(A).

Look them up on www.mathworld.wolfram.com.

I'm not going to do it all for you.

zaza

[krtxmrtz]

Do you know how to differentiate by first principles?

Even so it's not so easy:

dy/dx = lim(h -> 0) [f(x+h) - f(x-h)]/2h

Applied to the sine:

[f(x+h) - f(x-h)]/2h = [sin(x+h) - sin(x-h)]/2h = [(sin(x)cos(h) + sin(h)cos(x) - sin(x)cos(h) + sin(h)cos(x)]/2h = cos(x)*sin(h)/h

So it remains to be proved that sin(h)/h -> 1 when h -> 0. This is something we all know and take for granted but the only straightforward way I know to prove it is by using the first 2 terms of the Taylor series for the sine. And as I said above, you can't do that as the Taylor series is derived using derivatives that you must know beforehand.

[zaza]

sin(h)/h -> 0

!

Ultimately, you could always go right back to the Taylor series, but somehow I think that that is not required here. Working off the back of something that has been proven is going to have to happen somewhere, and I suspect going so far as to get into the nitty gritty of Taylor series in order to prove this is going a bit far.

zaza

[krtxmrtz]

!

zaza

I meant 1, of course and I've fixed my previous post already.

Ultimately, you could always go right back to the Taylor series, but somehow I think that that is not required here. Working off the back of something that has been proven is going to have to happen somewhere, and I suspect going so far as to get into the nitty gritty of Taylor series in order to prove this is going a bit far.
zaza

I agree, but I think this one needs some trick that's not in my big bag. When I used to study calculus, we worked with a concept called infinitessimals or something like that, they were pairs of functions that had approximately the same value when the variable tended to something. x and sin(x) formed such a pair for x->0, i.e. sin(x) = x (approx) as x->0 but I can't remember if/how this could be proved.

[zaza]

...they were pairs of functions that had approximately the same value when the variable tended to something. x and sin(x) formed such a pair for x->0, i.e. sin(x) = x (approx)

I thought that was what you meant by "Taylor Series". The expansion of sinx is:

x-(x^3)/3!+(x^5)/5!-...

which for small x leads to the approximation sinx~x. Of course, at 0 you'd end up with 0/0 so you use L'Hopital's rule to determine the value - which gives you cosx/1 = 1.

zaza

[krtxmrtz]

I have found an elegant proof for:

lim (h -> 0) sin(h)/h = 1

Because sin(-h) / -h = sin(h) / h then it is sufficient to consider lim(h -> 0+) sin(h) / h (though in the following, for convenience I'll write 0 rather than 0+)

In the attached figure let h be a small positive angle.

It can be assumed without loss of generality that segment OA = OB = 1 and from this:
OC = cos(h) and BC = sin(h)

The following inequalities between areas hold:

Sector(OCD) <= Tringle(OBC) <= Sector(OBA)

In terms of h:

(1/2)*h*cos²(h) <= (1/2)*sin(h)*cos(h) <= h/2

Dividing everything by (1/2)*h*cos(h), which is a positive quantity because h is positive:

cos(h) <= sin(h) / h <= 1/cos(h)

When h -> 0, cos(h) -> 1 and 1/cos(h) -> 1 so:

1 <= lim(h -> 0) sin(h) / h <= 1

hence, lim(h -> 0) sin(h) / h = 1

[tedz0r]

thanks zaza and ktrzmahlaukfhlsuth for ur help, i finally got it, after ages.. couldnt hav don it without ur posts, still heaps for me to learn though

[krtxmrtz]

still heaps for me to learn though

Same here