How do you find the derivative of ##y=cos(x)## from first principle?

Using the definition of a derivative:

##dy/dx = lim_(h->0) (f(x+h)-f(x))/h##, where ##h = deltax##

We substitute in our function to get:

##lim_(h->0) (cos(x+h)-cos(x))/h##

Using the Trig identity:

##cos(a+b) = cosacosb - sinasinb##,

we get:

##lim_(h->0) ((cosxcos h - sinxsin h)-cosx)/h##

Factoring out the ##cosx## term, we get:

##lim_(h->0) (cosx(cos h-1) - sinxsin h)/h##

This can be split into 2 fractions:

##lim_(h->0) (cosx(cos h-1))/h - (sinxsin h)/h##

Now comes the more difficult part: recognizing known formulas.

The 2 which will be useful here are:

##lim_(x->0) sinx/x = 1##, and ##lim_(x->0) (cosx-1)/x = 0##

Since those identities rely on the variable inside the functions being the same as the one used in the ##lim## portion, we can only use these identities on terms using ##h##, since that's what our ##lim## uses. To work these into our equation, we first need to split our function up a bit more:

##lim_(h->0) (cosx(cos h-1))/h - (sinxsin h)/h##

becomes:

##lim_(h->0)cosx((cos h-1)/h) - sinx((sin h)/h)##

Using the previously recognized formulas, we now have:

##lim_(h->0) cosx(0) - sinx(1)##

which equals:

##lim_(h->0) (-sinx)##

Since there are no more ##h## variables, we can just drop the ##lim_(h->0)##, giving us a final answer of: ##-sinx##.