What is the derivative of #-sin(x)#?

Question

Scarlett Allison · Accepted Answer

Here is the correct derivation; the previous answer has errors. First of all, the minus sign in front of a function #f(x)=-sin(x)#, when taking a derivative, would change the sign of a derivative of a function #f(x)=sin(x)# to an opposite. This is an easy theorem in the theory of limits: limit of a constant multiplied by a variable equals to this constant multiplied by a limit of a variable. So, let's find the derivative of #f(x)=sin(x)# and then multiply it by #-1#. We have to start from the following statement about the limit of trigonometric function #f(x)=sin(x)# as its argument tends to zero:
#lim_(h->0)sin(h)/h=1# Proof of this is purely geometrical and is based on a definition of a function #sin(x)#. There are many Web resources that contain a proof of this statement, like The Math Page. Using this, we can calculate a derivative of #f(x)=sin(x)#:
#f'(x)=lim_(h->0) (sin(x+h)-sin(x))/h#
Using representation of a difference of #sin# functions as a product of #sin# and #cos# (see Unizor , Trigonometry - Trig Sum of Angles - Problems 4) ,
#f'(x)=lim_(h->0) (2*sin(h/2)cos(x+h/2))/h#
#f'(x)=lim_(h->0) sin(h/2)/(h/2)*lim_(h->0)cos(x+h/2)#
#f'(x)=1*cos(x)=cos(x)# Therefore, derivative of #f(x)=-sin(x)# is #f'(x)=-cos(x)#.

Scarlett Allison · Answer

undefined The derivative of -sin(x) is -cos(x).