Find the derivative of the function: sec x | vedclass

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(A) Let $f(x) = \sec x$. Accordingly,from the first principle of derivatives:$f'(x) = \lim_{h 	o 0} \frac{f(x+h) - f(x)}{h}$$= \lim_{h 	o 0} \frac{\

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$\sec x 	an x$

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(A) Let $f(x) = \sec x$. Accordingly,from the first principle of derivatives:$f'(x) = \lim_{h 	o 0} \frac{f(x+h) - f(x)}{h}$$= \lim_{h 	o 0} \frac{\sec(x+h) - \sec x}{h}$$= \lim_{h 	o 0} \frac{1}{h} \left[ \frac{1}{\cos(x+h)} - \frac{1}{\cos x} ight]$$= \lim_{h 	o 0} \frac{1}{h} \left[ \frac{\cos x - \cos(x+h)}{\cos x \cos(x+h)} ight]$Using the identity $\cos A - \cos B = -2 \sin \left( \frac{A+B}{2} ight) \sin \left( \frac{A-B}{2} ight)$:$= \frac{1}{\cos x} \cdot \lim_{h 	o 0} \frac{-2 \sin \left( \frac{2x+h}{2} ight) \sin \left( \frac{-h}{2} ight)}{h \cos(x+h)}$$= \frac{1}{\cos x} \cdot \lim_{h 	o 0} \left[ \frac{\sin \left( \frac{2x+h}{2} ight)}{\cos(x+h)} \cdot \frac{\sin(h/2)}{h/2} ight]$$= \frac{1}{\cos x} \cdot \frac{\sin x}{\cos x} \cdot 1$$= \sec x 	an x$

Find the derivative of the function: $\sec x$

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