hi,
this is sajith. iam confused. I want some body to explain me in detail about derivatives.
HOW , WHEN Y=SINX

dY/dX=COSX ?

WHAT DO U MEAN BY CONTINUITY OF A FUNCTION? WHEN WILL U SAY A FUNCTION TO BE CONTINUOUS.
HOW THE CONTINUITY RELATED WITH DIFFERENTIABILITY?


PLS , ANYBODY SENT AN ELABORATE EXPLANATION.

Ok, its been a while since I have done calculus from first principles, but I will give it a try. We know that the derivative of a function is the slope of the tangent at the point in question. And to approximate the slope we can take a secant using that point and another point either to the left or the right of it. (Newton's forward and backward methods) And to get a better approximation of the slope of the tangent we just make the difference between the two points smaller. So if we were looking for the tangent of a function at x=4 we could use the other point at say x=3 to get one approximation of the slope of the tangent. A better approximation would be using the point x=3.5. An even better approximation would be using x=3.99. If we shrink that distance to zero (bringing the to points to one point) we theoretically get the tangent of the line. But we only looked at it approaching the point from the left. We need to also approach it from the right. That is why continuity is important. Because if a function is discontinous at a point, you can't approach some points from both sides.

Now, there are plenty of ways you can show that d(sinx)/dx=cosx. One would be to use first principles (ie, limits), or another would be to use the infinite series definitions of the trig functions. Then you can take the derivatives of a polynomial function and see that the derivative of sinx is cosx.
Let me TRY to go through the first principles way now.
(when I type lim, I am just shortening the phrase 'the limit as h goes to zero').

So we know first principles is

lim [f(x+h)-f(x)]/h

is the derivative. So lets use f(x)=sinx, so we get


lim[sin(x+h)-sin(x)]/h=lim[sin(x)cos(h)+cos(x)sin(h)-sin(x)]/h

After this point its about knowing how to manipulate this to get rid of the h in the denominator so we don't divide by zero.


lim{[sinxcosh-sinx]/h+cosxsinh/h},

then recognizing the two identities

lim[(cosh-1)/h]=0, and lim[sinh/h]=1 we can write the above equation as

lim[sinx]*lim[(cosh-1)/h]+lim[cosx]lim[sinh/h]


and since lim[cosx] (read as the limit of cosine x as h goes to zero) is cosx, and the other term is zero we get that d(sinx)/dx=cosx.
A similar procedure will show that d(cosx)/dx=-sinx.
Hope this helps

Oh. I didn't realize derivatives were so simple!

I understand your proof, but I have to proove this dervitive with no identities. It is possible for me to use identities, only if I proove them aswell. I would appreciate it if you could proove the two identities in your post.


The identities ->

lim[(cosh-1)/h]=0
lim[sinh/h]=1


If you could help, I'd appreciate it. I'm having problems, this is boggling my mind.

Continous means the function is not split into pieces. A rather crude way of describing it would be to say "You can draw the function without taking your pen off the paper". If a function is differentiable then it is also continous. You can see this by the definition of differentiable being well defined,

f'(x) = [f(x+h)-f(x)]/h

If the limit as h->0 of f(x+h) is NOT f(x) (ie the function is not continous) then you will get a non-zero amount in the numerator and a zero amount in the denominator at h=0, which is not defined.

Think about the Step function, which is

f(x) = 0 if x<0
f(x) = 1 if x>0

At x=0 you have a problem, it's not continous. You can see that the gradient (ie value of the derivative) of the function at x=0 is infinite, it's not well defined. Hence you MUST have continuity for a derivative to exist. You must also have another thing, since continuity is nessacery but not sufficent for differentiability.

The analysis way of doing sin's derivative is to define sin and cos in terms of power series and from there it follows immediately their derivatives relate to one another along with all their other properties.

The 'old' way of showing (sin h)/h -> 1 was to do it geometrically by drawing triangles. Every time I've seen a question asking for you to prove (d/dx)sin x = cos x it's always said "You may assume sin h/h -> 1 as h->0", if not then you have to do WAY more work than you should.

I agree with "Guest_Mike", can anyone write a proof, proving them two identities?

http://oregonstate.edu/instruct/mth251/cq/...esson/sine.html

They do it there and say that the sinh/h ->1 and cosh-1/h ->0 is done elsewhere in Stage 3 and 4. I'd find the exact pages but I have to head out now. Shouldn't be too hard to find if you're desperate

A hint that lim x->0 (sin x)/x = 1 .... http://oregonstate.edu/instruct/mth251/cq/...sinExample.html

A proof of the above statement:
http://oregonstate.edu/instruct/mth251/cq/...on/squeeze.html
http://oregonstate.edu/instruct/mth251/cq/...n/sinProof.html
http://oregonstate.edu/instruct/mth251/cq/...sinProof.1.html