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