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प्रश्न
Differentiate each of the following from first principle:
\[\frac{\cos x}{x}\]
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उत्तर
\[ \frac{d}{dx}\left( f(x) \right) = \lim_{h \to 0} \frac{f\left( x + h \right) - f\left( x \right)}{h}\]
\[ = \lim_{h \to 0} \frac{\frac{\cos \left( x + h \right)}{x + h} - \frac{\cos x}{x}}{h}\]
\[ = \lim_{h \to 0} \frac{x \cos \left( x + h \right) - \left( x + h \right) \cos x}{h x \left( x + h \right)}\]
\[ = \lim_{h \to 0} \frac{x \left( \cos x \cos h - \sin x \sin h \right) - x \cos x - h \cos x}{h x \left( x + h \right)}\]
\[ = \lim_{h \to 0} \frac{x \cos x \cos h - x \sin x \sin h - x \cos x - h \cos x}{h x \left( x + h \right)}\]
\[ = \lim_{h \to 0} \frac{x \cos x \cos h - x \cos x - x \sin x \sin h - h \cos x}{h x \left( x + h \right)}\]
\[ = x\cos x \lim_{h \to 0} \frac{\cos h - 1}{h} - \frac{x\sin x}{x} \lim_{h \to 0} \frac{\sin h}{h} \lim_{h \to 0} \frac{1}{x + h} - \frac{\cos x}{x} \lim_{h \to 0} \frac{1}{x + h}\]
\[ = x \cos x \lim_{h \to 0} \frac{- 2 \sin^2 \frac{h}{2}}{\frac{h^2}{4}} \times \frac{h}{4} - \frac{x\sin x}{x} \lim_{h \to 0} \frac{\sin h}{h} \lim_{h \to 0} \frac{1}{x + h} - \frac{\cos x}{x} \lim_{h \to 0} \frac{1}{x + h} \left[ \because \lim_{h \to 0} \frac{\sin^2 \frac{h}{2}}{\frac{h^2}{4}} = \lim_{h \to 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}} \times \lim_{h \to 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}} = 1 \times 1, i . e . 1 \right]\]
\[ = - x \cos x \lim_{h \to 0} \frac{h}{2} - \frac{x\sin x}{x} \lim_{h \to 0} \frac{\sin h}{h} \lim_{h \to 0} \frac{1}{x + h} - \frac{\cos x}{x} \lim_{h \to 0} \frac{1}{x + h}\]
\[ = - x \cos x \times 0 - \sin x \left( 1 \right)\frac{1}{x} - \frac{\cos x}{x}\frac{1}{x}\]
\[ = 0 - \frac{\sin x}{x} - \frac{\cos x}{x^2}\]
\[ = - \frac{\sin x}{x} - \frac{\cos x}{x^2}\]
\[ = \frac{- x \sin x - \cos x}{x^2}\]
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