Find the derivative of cos using First Principles?

1 Answer
Jul 18, 2017

By the limit definition of the derivative if y=f(x), then

dy/dx=f'(x) = lim_(h rarr 0) ( f(x+h)-f(x) ) / h

So with y=f(x) = cosx we have;

f'(x) = lim_(h rarr 0) ( cos(x+h) - cos x ) / h

Using the cosine sum of angle formula:

cos(A+B)=cosAcosB- sinAcosB

We get

f'(x) = lim_(h rarr 0) ( cosxcos h-sinxsin h - cos x ) / h
" " = lim_(h rarr 0) ( cosxcos h-cosx-sinxsin h ) / h
" " = lim_(h rarr 0) ( cosx(cos h-1)-sinxsin h ) / h
" " = lim_(h rarr 0) {(cosx(cos h-1))/h-(sinxsin h ) / h}
" " = lim_(h rarr 0) (cosx(cos h-1))/h- lim_(h rarr 0)(sinxsin h ) / h
" " = cosxlim_(h rarr 0) (cos h-1)/h - sinx lim_(h rarr 0)(sin h ) / h

We now have to rely on some standard calculus limits:

lim_(h rarr 0)sin h/h =1
lim_(h rarr 0)(cos h-1)/h =0

And so using these we have:

f'(x) = cosx(0) - sinx (1)
" " = - sinx

Hence,

dy/dx=-sinx