Limits and Derivatives

Question

Derivative of sin x :

Answer

Let    $space space space space space space straight f left parenthesis straight x right parenthesis space equals space space sin space straight x comma space Therefore comma space straight f left parenthesis straight x right parenthesis space space equals space limit as straight h rightwards arrow 0 of fraction numerator straight f left parenthesis straight x plus straight h right parenthesis minus straight f left parenthesis straight x right parenthesis over denominator straight h end fraction$
$space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space limit as straight h rightwards arrow 0 of fraction numerator sin left parenthesis straight x plus straight h right parenthesis minus sinx over denominator straight h end fraction equals limit as straight h rightwards arrow 0 of fraction numerator 2 cos open parentheses begin display style fraction numerator straight x plus straight h plus straight x over denominator 2 end fraction end style close parentheses sin open parentheses begin display style fraction numerator straight x plus straight h minus straight x over denominator 2 end fraction end style close parentheses over denominator straight h end fraction$
$space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space limit as straight h rightwards arrow 0 of fraction numerator 2 cos open parentheses straight x straight h over 2 close parentheses sin space begin display style straight h over 2 end style over denominator straight h end fraction equals space limit as straight h rightwards arrow 0 space cos space open parentheses straight x straight h over 2 close parentheses. limit as straight h over 2 rightwards arrow 0 of of fraction numerator sin begin display style straight h over 2 end style over denominator begin display style straight h over 2 end style end fraction$

= cos (x+0).1 = cos x    $space space space space space space space space space space open square brackets because limit as straight x rightwards arrow 0 of fraction numerator sin space straight x over denominator straight x end fraction equals 1 close square brackets$
So, $space space space space space space space space space space space space space space space space space space space space space space space space straight d over dx left parenthesis sinx right parenthesis space equals space xos space straight x space for space all space straight x element of straight R$ o,