prove the derivative of cosx=-sinx using the definition of a derivative?

Answer 1

Quoting my answer to the nearly identical question here: http://answers.yahoo.com/question/index;_ylt=Al.7r9bsOnRzG_CIpn1Shvfsy6IX?qid=20070603125901AAKtIfQ&show=7#profile-info-d04094d055d90c1fd655614bf175cb61aa

[h→0]lim (cos (x+h) - cos x)/h

Using the angle addition formula:

[h→0]lim (cos x cos h - sin x sin h - cos x)/h

Splitting the fraction:

[h→0]lim ((cos h - 1)/h) cos x - (sin h/h) sin x

Using the linearity of limits:

cos x [h→0]lim (cos h - 1)/h - sin x [h→0]lim (sin h/h)

[h→0]lim (sin h/h) = 1 (this is proved geometrically, before even considering the derivatives), so:

cos x [h→0]lim (cos h - 1)/h - sin x

Multiplying the second limit by the conjugate:

cos x [h→0]lim (cos² h - 1)/((cos h + 1)h) - sin x

Using the Pythagorean theorem:

cos x [h→0]lim -sin² h/((cos h + 1)h) - sin x

Writing as the product of two fractions:

cos x [h→0]lim -sin h/(cos h + 1) * sin h/h - sin x

The right fraction approaches 1 (as I mentioned earlier), and the left fraction can be evaluated directly as 0/2, so we have:

cos x * 0 * 1 - sin x

-sin x

And we are done.

Answer 2

lim (f(x + dx) - f(x))/dx = (cos(x + dx) - cosx)/dx =
dx->0
= (-2*sin(2*x + dx)/2*sin(dx/2))/dx = (-2*sin(2*x + dx)/2*sin(dx/2))/((dx/2)*2) = -sin((2*x + dx)/2) = -sin(x)

<0>>
It would be more clear when you write it down!

Answer 3

This has been done here:

http://au.answers.yahoo.com/question/index;_ylt=AjJFWGLPV.0Yx1khz2iDYz3h5gt.?qid=20070527150303AAvVlTO&show=7#profile-info-EsNDPTxhaa

Have a look and decide

Answer 4

Let y = f(x) = cos(x)
At a small interval dx,

y + dy = cos(x+dx).

Now derivative is defined as :

limit dx-> 0 (dy/dx) = f'(x)

ie, (cos(x+dx) - cos(x)) / dx

cos(x+dx) can be written as cos(x).cos(dx) - sin(x).sin(dx)

Therefore,

dy/dx =

cos(x).cos(dx) - sin(x).sin(dx) - cos(x)
-----------------------------------------------------
dx

= cos(x). (cos(dx) -1) - sin(x). sin(dx)
--------------- --------------------
dx dx

Applying the limit dx -> 0, we have,
limit dx->0 cos(dx) -1 cos(dx) 1
--------------- = ---------- - ----
dx dx dx

= 1/0 - 1/0
= 0

And,

limit dx->0 sin(dx)/dx = 1

Putting these , we have the derivative as,

=====================================
f'(x) = cos(x) . (0) - sin(x). (1)
= - sin(x)
=====================================