Question

Find the derivative of `cos` using First Principles?

Answer 1

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`