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प्रश्न
Differentiate each of the following from first principle:
x2 sin x
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उत्तर
\[ \frac{d}{dx}\left( f(x) \right) = \lim_{h \to 0} \frac{f\left( x + h \right) - f\left( x \right)}{h}\]
\[ = \lim_{h \to 0} \frac{\left( x + h \right)^2 \sin \left( x + h \right) - x^2 \sin x}{h}\]
\[ = \lim_{h \to 0} \frac{\left( x^2 + h^2 + 2xh \right)\left( \sin x \cos h + \cos x \sin h \right) - x^2 \sin x}{h}\]
\[ = \lim_{h \to 0} \frac{x^2 \sin x \cos h + x^2 \cos x \sin h + h^2 \sin x \cos h + h^2 \cos x \sin h + 2xh \sin x \cos h + 2xh \cos x \sin h - x^2 \sin x}{h}\]
\[ = \lim_{h \to 0} \frac{x^2 \sin x \cos h - x^2 \sin x + x^2 \cos x \sin h + h^2 \sin x \cos h + h^2 \cos x \sin h + 2xh \sin x \cos h + 2xh \cos x \sin h}{h}\]
\[ = x^2 \sin x \lim_{h \to 0} \frac{\cos h - 1}{h} + x^2 \cos x \lim_{h \to 0} \frac{\sin h}{h} + \sin x \lim_{h \to 0} h \cos h + \cos x \lim_{h \to 0} h \sin h + 2x \sin x \lim_{h \to 0} \cosh + 2x \cos x \lim_{h \to 0} \sin h\]
\[ = x^2 \sin x \lim_{h \to 0} \frac{- 2 \sin^2 \frac{h}{2}}{\frac{h^2}{4}} \times \frac{h}{4} + x^2 \cos x \lim_{h \to 0} \frac{\sin h}{h} + \sin x \lim_{h \to 0} h \cos h + \cos x \lim_{h \to 0} h \sin h + 2x \sin x \lim_{h \to 0} \cosh + 2x \cos x \lim_{h \to 0} \sin h \left[ \because \lim_{h \to 0} \frac{\sin^2 \frac{h}{2}}{\frac{h^2}{4}} = \lim_{h \to 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}} \times \lim_{h \to 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}} = 1 \times 1, i . e . 1 \right]\]
\[ = - x^2 \sin x \times \lim_{h \to 0} \frac{h}{2} + x^2 \cos x \lim_{h \to 0} \frac{\sin h}{h} + \sin x \lim_{h \to 0} h \cos h + \cos x \lim_{h \to 0} h \sin h + 2x \sin x \lim_{h \to 0} \cosh + 2x \cos x \lim_{h \to 0} \sin h \]
\[ = - x^2 \sin x \times 0 + x^2 \cos x \left( 1 \right) + \sin x \left( 0 \right) + \cos x \left( 0 \right) + 2x \sin x \left( 1 \right) + 2x \cos x \left( 0 \right)\]
\[ = 0 + x^2 \cos x + 2x \sin x\]
\[ = 0 + x^2 \cos x + 2x \sin x\]
\[ = x^2 \cos x + 2x \sin x\]
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