Question

Find the derivative of `sinx` using First Principles?

Answer 1

This is a Topic at Socratic. See the various posts here: https://socratic.org/calculus/differentiating-trigonometric-functions/differentiating-sinx-from-first-principles

Answer 2

By definition of the derivative `f'(x)=lim_(h rarr 0) ( f(x+h)-f(x) ) / h`
So with `f(x) = sinx` we have;

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

Using `sin (A+B)=sinAcosB+sinBcosA` we get

`f'(x)=lim_(h rarr 0) ( sinxcos h+sin hcosx - sin x ) / h`
`f'(x)=lim_(h rarr 0) ( sinx(cos h-1)+sin hcosx ) / h`
`f'(x)=lim_(h rarr 0) ( (sinx(cos h-1))/h+(sin hcosx) / h )`

`f'(x)=lim_(h rarr 0) (sinx(cos h-1))/h+lim_(h rarr 0)(sin hcosx) / h`

`f'(x)=(sinx)lim_(h rarr 0) (cos h-1)/h+(cosx)lim_(h rarr 0)(sin h) / h`

We know have to rely on some standard 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)=0+(cosx)(1) =cosx`

Hence,

`d/dxsinx=cosx`