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Question Number 192867 by beto last updated on 30/May/23

$$derivateofcsc\left(2x\right)bydefinition$$

Answered by cortano12 last updated on 02/Jun/23

$$\frac{\mathrm{d}\left(\frac{1}{\sin 2\mathrm{x}}\right)}{\mathrm{dx}}=\underset{\mathrm{h}\to 0}{\lim }\frac{\frac{1}{\sin (2\mathrm{x}+2\mathrm{h})}-\frac{1}{\sin 2\mathrm{x}}}{\mathrm{h}}$$$$=\underset{\mathrm{h}\to 0}{\lim }\frac{\sin 2\mathrm{x}-\sin (2\mathrm{x}+2\mathrm{h})}{\mathrm{h}\sin 2\mathrm{x}\sin (2\mathrm{x}+2\mathrm{h})}$$$$=\frac{1}{{\sin }^{2}2\mathrm{x}}\underset{\mathrm{h}\to 0}{\lim }\frac{2\cos (2\mathrm{x}+\mathrm{h})\sin (-\mathrm{h})}{\mathrm{h}}$$$$=\frac{-2\cos 2\mathrm{x}}{\sin {}^{2}2\mathrm{x}}=-2\cot 2\mathrm{x}\csc 2\mathrm{x}$$

Commented by Subhi last updated on 30/May/23

$$itmustbe-2cot\left(2x\right).csc\left(2x\right)$$$$As$$$$sin\left(2x\right)-sin(2x+2h)=-2cos(2x+h)sin\left(h\right)$$

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