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