Let us use the basic rule of differentiation
#[f(x)]'=lim_(hrarr0) (f(x+h)-f(x))/h#
Applying this rule:
#[cos(x)]′=lim_(hrarr0) (cos(x+h)-cos(x))/h#
Using the angle sum identity:
#cos(x+y)=cos(x)cos(y)-sin(x)sin(y)#
We get:
#[cos(x)]′=lim_(hrarr0) (cos(x)cos(h)-sin(x)sin(h)-cos(x))/h#
#=lim_(hrarr0) (cos(x)cos(h)-cos(x)-sin(x)sin(h))/h#
Taking #cosx# common from first two terms we get:
#[cos(x)]′=lim_(hrarr0) (cos(x)(cos(h)-1)-sin(x)sin(h))/h#
#=lim_(hrarr0) (cos(x)(cos(h)-1))/h-lim_(hrarr0) (sin(x)sin(h))/h#
#=cos(x)lim_(hrarr0) (cos(h)-1)/h-sin(x)lim_(hrarr0) sin(h)/h#
Taking each two terms alone, as #hrarr0, cos(h)rarr1 and sin(h)rarr0#
#lim_(hrarr0) (cos(h)-1)/h=lim_(hrarr0) (1-1)/h=0#
#lim_(hrarr0) sin(h)/h=1# since both values are going to zero but do not reach zero
#[cos(x)]′=cos(x)lim_(hrarr0) (cos(h)-1)/h-lim_sin(x)(hrarr0)sin(h)/h#
#=cos(x)(0)-sin(x)(1)#
#=-sinx#
