Find the derivative of cos x from first princ | vedclass

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Question

(A) Let $f(x) = \cos x$. According to the first principle of derivatives:$f'(x) = \lim_{h 	o 0} \frac{f(x+h) - f(x)}{h}$$= \lim_{h 	o 0} \frac{\cos(

Vedclass · Accepted Answer

$-\sin x$

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(A) Let $f(x) = \cos x$. According to the first principle of derivatives:$f'(x) = \lim_{h 	o 0} \frac{f(x+h) - f(x)}{h}$$= \lim_{h 	o 0} \frac{\cos(x+h) - \cos x}{h}$Using the identity $\cos(A+B) = \cos A \cos B - \sin A \sin B$:$= \lim_{h 	o 0} \frac{\cos x \cos h - \sin x \sin h - \cos x}{h}$$= \lim_{h 	o 0} \left[ \cos x \left( \frac{\cos h - 1}{h} ight) - \sin x \left( \frac{\sin h}{h} ight) ight]$$= \cos x \cdot (0) - \sin x \cdot (1)$$= -\sin x$

Find the derivative of $\cos x$ from first principle.

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