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Question
Differentiate of the following from first principle:
sin (2x − 3)
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Solution
\[\frac{d}{dx}\left( f(x) \right) = \lim_{h \to 0} \frac{f\left( x + h \right) - f\left( x \right)}{h}\]
\[ = \lim_{h \to 0} \frac{\sin \left( 2x + 2h - 3 \right) - \sin \left( 2x - 3 \right)}{h}\]
\[\text{ We know }:\]
\[sin C-sin D=2 cos\left( \frac{C + D}{2} \right)\sin\left( \frac{C - D}{2} \right)\]
\[ = \lim_{h \to 0} \frac{2 \cos \left( \frac{2x + 2h - 3 + 2x - 3}{2} \right) \sin \left( \frac{2x + 2h - 3 + 2x - 3}{2} \right)}{h}\]
\[ = \lim_{h \to 0} \frac{2 \cos \left( \frac{4x + 2h - 6}{2} \right) \sin \left( h \right)}{h}\]
\[ = \lim_{h \to 0} 2 \cos \left( \frac{4x + 2h - 6}{2} \right) \lim_{h \to 0} \frac{\sin h}{h}\]
\[ = 2 \cos \left( \frac{4x - 6}{2} \right) \left( 1 \right)\]
\[ = 2 \cos \left( 2x - 3 \right)\]
\[ \]
\[\]
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