Cos Derivative Proof

[Jason76]

\(\displaystyle f(x) = \cos(x)\)

\(\displaystyle f'(x) = \dfrac{d}{dx} \cos(x) = -\sin(x)\)

Proof

Using: \(\displaystyle \lim h \to 0[\dfrac{(x + h) - (f(x))}{h}]\)

\(\displaystyle \lim h \to 0[\dfrac{\cos(x + h) - (\cos(x))}{h}]\)

\(\displaystyle \lim h \to 0[\dfrac{[\cos(x)\cos(h) - \sin(x)\sin(h)] - (\sin(x))}{h}]\) using sum and difference identity \(\displaystyle \cos(A + B) = \cos(A) \cos(B) - \sin(A) \sin(B)\) for \(\displaystyle \cos(x + h)\) Next move

 

[fcabanski]

I have to learn how to use the latex code.

Your third step is not correct. You changed cos(x) into sin(x). The correct equation is:

cos(x)cos(h) - sin(x)sin(h) - cos(x) --- it's all over h, but let's just deal with the numerator.

Rewrite: cos(x)cos(h) - cos(x) - sin(x)sin(h)

Factor out cos(x): cos(x)[cos(h)-1] - sin(x)sin(h)

That is: (cos(x)[cos(h)-1] - sin(x)sin(h) )/h =

Separate into two fractions: (cos(x)[cos(h)-1])/h - sin(x)sin(h)/h

Recall that as h ---> 0 (cos(h)-1)/h = 0 and sin(h)/h = 1.

cos(x)*0 - sin(x)*1 = -sin(x)

 

[Deleted member 4993]

It is very similar to:

http://www.freemathhelp.com/forum/threads/83611-Sin-Derivative-Proof