Solution to a problem requested by one student

Prove d/dx cos x = - sin x from first principles

d/dx cos x = lim δx→0 [ { f(x + δx ) - f(x) } / δx ]

= lim δx →0 [{ cos ( x + δx ) - cos x } / δx ].

[ cos A - cos B ] = - 2 [ sin (A + B)/2 • sin (A - B)/2 ]

= lim δx →0 (- 2 ) [ sin ( x + δx + x ) / 2 • sin ( x + δx - x )/ 2 ] / δx

= lim δx →0 (- 2 ) [ sin ( 2x + δx ) /2 • sin δx /2 ] / δx

= lim δx →0 (- 2 ) [  sin( 2x / 2 + δx / 2 ) • sin δx / 2 ] / δx

[ bring ( 2 ) from numerator to the denominator ]

= lim δx → 0 ( - ) [ sin ( x + δx / 2 ) • sin δx /2 ] /δx /( 2 )

= lim δx → 0 ( - ) [ sin ( x + δx / 2 ) • sin ( δx /2 )/( δx/2 ) ]

= ( - sin x ) - { 0 • ( 1 ) }  

= ( - sin x ) - ( 0 ) = ( - sin x )

[ lim δx → 0 ( sin δx / δx ) = 1, lim δx → 0 ( sin x ) = sin x , lim δx → 0 ( δx/2 ) = 0/2 = 0 ]

d/dx ( cos x ) = ( - sin x  )        hence proved