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

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Nov 3, 2016

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#.

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