Let f(x) =\( cos(x-\dfrac{\pi}{8})\) Accordingly, f(x + h) = cos 8

Accordingly, f(x + h)

\( cos(x+h-\dfrac{\pi}{8})\)

By first principle,

\( f'(x)=\lim\limits_{h \to 0}\dfrac{f(x+h)-f(x)}{h}\)

\( \lim\limits_{h \to 0} \dfrac{1}{h}[cos(x+h-\dfrac{\pi}{8})\)

\(-cos(\pi-\dfrac{\pi}{8})]\)

\( \lim\limits_{h \to 0} \dfrac{1}{h}[-2sin(\dfrac{2x+h-\dfrac{\pi}{4}}{2})sin(\dfrac{h}{2})]\)

\([-sin(\dfrac{2x+h-\dfrac{\pi}{4})}{2}\dfrac{sin(\dfrac{h}{2})}{(\dfrac{h}{2})}]\)

= \(-sin( \dfrac{2x+0-\dfrac{\pi}{4}}{2}).1\)

= \( sin(x-\dfrac{\pi}{8})\)

Answered by Pragya Singh | 1 year ago