Let, f(x) = sin(x + a) f(x + h) = sin(x + h + a)
By first principle,
\( f'(x)=\lim\limits_{h \to 0}\dfrac{f(x+h)-f(x)}{h} \)
\(\lim\limits_{h \to 0}\dfrac{sin(x+h+a)-sin(x+a)}{h} \)
\(\lim\limits_{h \to 0}\dfrac{1}{h}[2cos(\dfrac{x+h+a+x+a}{2})\)
\( sin(\dfrac{x+h+a-x-a}{2})]\)
\( \lim\limits_{h \to 0}\dfrac{1}{h}[2cos(\dfrac{2x+2a+h}{2})sin(\dfrac{h}{2})]\)
\( \lim\limits_{h \to 0}[cos(\dfrac{2x+2a+h}{2})[\dfrac{sin(\dfrac{h}{2})}{(\dfrac{h}{2})}]]\)
\( \lim\limits_{h \to 0}cos(\dfrac{2x+2a+h}{2}). \lim\limits_{\dfrac{h}{2} \to 0}[\dfrac{sin(\dfrac{h}{2})}{(\dfrac{h}{2})}]\)
\(cos( \dfrac{2x+2a}{2})\times 1\)
=cos (x + a)
Answered by Abhisek | 1 year ago