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