Limits and Derivatives — Maths STD 11 Science — Question

$f(x) = \cos \left( {x - \frac{\pi }{8}} \right)$
Then f (x + h) = $\cos \left( {x + h - \frac{\pi }{8}} \right)$
We know that $f{\text{'}}(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h}$
$\Rightarrow f{\text{'}}(x) = \mathop {\lim }\limits_{h \to 0} \frac{{\cos \left( {x + h - \frac{\pi }{8}} \right) - \cos \left( {x - \frac{\pi }{8}} \right)}}{h}$
$= \mathop {\lim }\limits_{h \to 0} \frac{{ - 2\sin \left( {x - \frac{\pi }{8} + \frac{h}{2}} \right)\sin \left( {\frac{h}{2}} \right)}}{h}$$= \mathop {\lim }\limits_{h \to 0} \frac{{ - \sin \left( {x - \frac{\pi }{8} + \frac{h}{2}} \right) \cdot \sin \left( {\frac{h}{2}} \right)}}{{\left( {\frac{h}{2}} \right)}}$
$= - \sin \left( {x - \frac{\pi }{8}} \right)$