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

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$-\operatorname{cosec} x \cot x$

Vedclass · Answer

(A) Let $f(x) = \operatorname{cosec} x$. Using the first principle of derivatives:$f'(x) = \lim_{h 	o 0} \frac{f(x+h) - f(x)}{h}$$= \lim_{h 	o 0} \frac{\operatorname{cosec}(x+h) - \operatorname{cosec} x}{h}$$= \lim_{h 	o 0} \frac{1}{h} \left[ \frac{1}{\sin(x+h)} - \frac{1}{\sin x} ight]$$= \lim_{h 	o 0} \frac{1}{h} \left[ \frac{\sin x - \sin(x+h)}{\sin x \sin(x+h)} ight]$Using the identity $\sin A - \sin B = 2 \cos(\frac{A+B}{2}) \sin(\frac{A-B}{2})$:$= \lim_{h 	o 0} \frac{1}{h} \left[ \frac{2 \cos(\frac{2x+h}{2}) \sin(-\frac{h}{2})}{\sin x \sin(x+h)} ight]$$= \lim_{h 	o 0} \left[ -\cos(\frac{2x+h}{2}) \cdot \frac{\sin(h/2)}{h/2} \cdot \frac{1}{\sin x \sin(x+h)} ight]$Since $\lim_{h 	o 0} \frac{\sin(h/2)}{h/2} = 1$:$= -\cos x \cdot 1 \cdot \frac{1}{\sin^2 x} = -\frac{\cos x}{\sin x} \cdot \frac{1}{\sin x} = -\cot x \operatorname{cosec} x$

Find the derivative of the function: $\operatorname{cosec} x$

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