Find the derivative of #sinx# using First Principles?
2 Answers
This is a Topic at Socratic. See the various posts here: https://socratic.org/calculus/differentiating-trigonometric-functions/differentiating-sinx-from-first-principles
By definition of the derivative
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 #
# 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#