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

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Using the limitation of a derivative:

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

We supply in our character 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 crack into 2 fractions:

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

Now comes the further involved part: recognizing known formulas.

The 2 which get be profitable short are:

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

Since those identities depend on the fickle internally the characters nature the corresponding as the one used in the ##lim## piece, we can singly use these identities on provisions using ##h##, gone that's what our ##lim## uses. To operation these into our equation, we pristine insufficiency to crack our character up a bit further:

##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 formal formulas, we now have:

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

which equals:

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

Since tshort are no further ##h## fickles, we can normal faint the ##lim_(h->0)##, giving us a terminal acceptance of: ##-sinx##.


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