show-that-derivative-of-Sin-x-x-1-

Question Number 150180 by jlewis last updated on 10/Aug/21

show that derivative of Sin x / x = 1

Answered by liberty last updated on 10/Aug/21

\frac{d}{dx} (\frac{\sin x}{x}) = \frac{xcos x - \sin x}{x^{2}}

Commented by liberty last updated on 10/Aug/21

y^{'} = \lim_{h \to 0} \frac{\frac{\sin (x + h)}{x + h} - \frac{\sin x}{x}}{h}

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

= \lim_{h \to 0} \frac{x (\sin xcos h + \cos xsin h) - (xsin x + hsin x)}{xh (x + h)}

= \lim_{h \to 0} \frac{xsin x (\cos h - 1) + xcos xsin h - hsin x}{xh (x + h)}

= \lim_{h \to 0} \frac{- 2 xsin x \sin^{2} h + xcos x \sin h - hsin x}{xh (x + h)}

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

= \frac{xcos x}{x^{2}} - \frac{\sin x}{x^{2}}

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