How do you differentiate #f(x)=sin(x)# from first principles?

1 Answer
Jun 12, 2018

# d/dxsinx=cosx#

Explanation:

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 #

# \ \ \ \ \ \ \ \ \=lim_(h rarr 0) ( sinx(cos h-1)+sin hcosx ) / h #

# \ \ \ \ \ \ \ \ \=lim_(h rarr 0) ( (sinx(cos h-1))/h+(sin hcosx) / h )#

# \ \ \ \ \ \ \ \ \=lim_(h rarr 0) (sinx(cos h-1))/h+lim_(h rarr 0)(sin hcosx) / h#

# \ \ \ \ \ \ \ \ \=(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 #, and # 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#