Find the derivative of the following function | vedclass

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Question

(A) Let $f(x) = \cos \left(x-\frac{\pi}{8}\right)$. Accordingly,$f(x+h) = \cos \left(x+h-\frac{\pi}{8}\right)$.By the first principle of derivatives:$

Vedclass · Accepted Answer

$-\sin \left(x-\frac{\pi}{8}\right)$

Vedclass · Answer

(A) Let $f(x) = \cos \left(x-\frac{\pi}{8}ight)$. Accordingly,$f(x+h) = \cos \left(x+h-\frac{\pi}{8}ight)$.By the first principle of derivatives:$f'(x) = \lim_{h 	o 0} \frac{f(x+h)-f(x)}{h}$$= \lim_{h 	o 0} \frac{1}{h} \left[ \cos \left(x+h-\frac{\pi}{8}ight) - \cos \left(x-\frac{\pi}{8}ight) ight]$Using the identity $\cos A - \cos B = -2 \sin \left(\frac{A+B}{2}ight) \sin \left(\frac{A-B}{2}ight)$:$= \lim_{h 	o 0} \frac{1}{h} \left[ -2 \sin \left(\frac{2x+h-\frac{\pi}{4}}{2}ight) \sin \left(\frac{h}{2}ight) ight]$$= \lim_{h 	o 0} \left[ -\sin \left(x + \frac{h}{2} - \frac{\pi}{8}ight) \cdot \frac{\sin(h/2)}{h/2} ight]$Since $\lim_{	heta 	o 0} \frac{\sin 	heta}{	heta} = 1$ as $h 	o 0 \Rightarrow \frac{h}{2} 	o 0$:$= -\sin \left(x - \frac{\pi}{8}ight) \cdot 1 = -\sin \left(x - \frac{\pi}{8}ight)$.

Find the derivative of the following function from first principles: $\cos \left(x-\frac{\pi}{8}\right)$

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