How do you differentiate #f(x)=sin(x)# from first principles?
1 Answer
# 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)=lim_(h rarr 0) ( sin(x+h) - sin x ) / h #
Using
# 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#