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प्रश्न
Differentiate each of the following from first principle:
\[\sqrt{\sin (3x + 1)}\]
उत्तर
\[ \frac{d}{dx}\left( f(x) \right) = \lim_{h \to 0} \frac{f\left( x + h \right) - f\left( x \right)}{h}\]
\[ = \lim_{h \to 0} \frac{\sqrt{\sin \left( 3\left( x + h \right) + 1 \right)} - \sqrt{\sin \left( 3x + 1 \right)}}{h} \]
\[ = \lim_{h \to 0} \frac{\sqrt{\sin \left( 3x + 3h + 1 \right)} - \sqrt{\sin \left( 3x + 1 \right)}}{h} \times \frac{\sqrt{\sin \left( 3x + 3h + 1 \right)} + \sqrt{\sin \left( 3x + 1 \right)}}{\sqrt{\sin \left( 3x + 3h + 1 \right)} + \sqrt{\sin \left( 3x + 1 \right)}}\]
\[ = \lim_{h \to 0} \frac{\sin \left( 3x + 3h + 1 \right) - \sin \left( 3x + 1 \right)}{h \left( \sqrt{\sin \left( 3x + 3h + 1 \right)} + \sqrt{\sin \left( 3x + 1 \right)} \right)}\]
\[We have:\]
\[ sin C-sin D= 2 cos\left( \frac{C + D}{2} \right)\sin\left( \frac{C - D}{2} \right)\]
\[ = \lim_{h \to 0} \frac{2 \cos \left( \frac{3x + 3h + 1 + 3x + 1}{2} \right) \sin \left( \frac{3x + 3h + 1 - 3x - 1}{2} \right)}{h \left( \sqrt{\sin \left( 3x + 3h + 1 \right)} + \sqrt{\sin \left( 3x + 1 \right)} \right)}\]
\[ = \lim_{h \to 0} \frac{2 \cos \left( \frac{6x + 3h + 2}{2} \right) \sin \frac{3h}{2}}{h \left( \sqrt{\sin \left( 3x + 3h + 1 \right)} + \sqrt{\sin \left( 3x + 1 \right)} \right)}\]
\[ = \lim_{h \to 0} 2 \cos \left( \frac{6x + 3h + 2}{2} \right) \lim_{h \to 0} \frac{\sin \frac{3h}{2}}{h \times \frac{3}{2}} \times \frac{3}{2} \times \lim_{h \to 0} \frac{1}{\left( \sqrt{\sin \left( 3x + 3h + 1 \right)} + \sqrt{\sin \left( 3x + 1 \right)} \right)} \]
\[ = 2 \cos \left( 3x + 1 \right) \times \left( \frac{3}{2} \right) \times \frac{1}{\sqrt{\sin \left( 3x + 1 \right)} + \sqrt{\sin \left( 3x + 1 \right)}}\]
\[ = \frac{3 \cos \left( 3x + 1 \right)}{2\sqrt{\sin \left( 3x + 1 \right)}}\]
\[ \]
\[\]
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