proof for the derivative of sin(x) is cos(x) (5 marks)
let f(x)=sin x f'(x) lim h-> 0 = ( sin(x+h) - sin(x))/h. f'(x) lim h-> 0 =( sin(x)cos(h) + cos(x)sin(h) - sin(x))/ h. f'(x) lim h-> 0=(sin(x)(cos(h)-1)/h + cos(x) (sin(h))/h. then as h tends to zero. (cos(h)-1)/h=0 and sin(h)/h =1. f'(x)= cos(x) QED
NP
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