Question

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

Answer 1

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`.