derivate-of-csc-2x-by-definition-

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

d e r i v a t e o f c s c (2 x) b y d e f i n i t i o n

Answered by cortano12 last updated on 02/Jun/23

\frac{d (\frac{1}{\sin 2 x})}{dx} = \lim_{h \to 0} \frac{\frac{1}{\sin (2 x + 2 h)} - \frac{1}{\sin 2 x}}{h}

= \lim_{h \to 0} \frac{\sin 2 x - \sin (2 x + 2 h)}{h \sin 2 x \sin (2 x + 2 h)}

= \frac{1}{\sin^{2} 2 x} \lim_{h \to 0} \frac{2 \cos (2 x + h) \sin (- h)}{h}

= \frac{- 2 \cos 2 x}{\sin^{2} 2 x} = - 2 \cot 2 x \csc 2 x

Commented by Subhi last updated on 30/May/23

i t m u s t b e - 2 c o t (2 x) . c s c (2 x)

A s

s i n (2 x) - s i n (2 x + 2 h) = - 2 c o s (2 x + h) s i n (h)

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