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प्रश्न
tan (2x + 1)
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उत्तर
\[\frac{d}{dx}\left( f(x) \right) = \lim_{h \to 0} \frac{f\left( x + h \right) - f\left( x \right)}{h}\]
\[ = \lim_{h \to 0} \frac{\tan \left( 2x + 2h + 1 \right) - \tan \left( 2x + 1 \right)}{h}\]
\[ = \lim_{h \to 0} \frac{\frac{sin \left( 2x + 2h + 1 \right)}{\cos \left( 2x + 2h + 1 \right)} - \frac{\sin \left( 2x + 1 \right)}{\cos \left( 2x + 1 \right)}}{h}\]
\[ = \lim_{h \to 0} \frac{sin \left( 2x + 2h + 1 \right) \cos \left( 2x + 1 \right) - \cos \left( 2x + 2h + 1 \right) \sin \left( 2x + 1 \right)}{h \cos \left( 2x + 2h + 1 \right) \cos \left( 2x + 1 \right)}\]
\[ = \lim_{h \to 0} \frac{\sin \left( 2x + 2h + 1 - 2x - 1 \right)}{h \cos \left( 2x + 2h + 1 \right) \cos \left( 2x + 1 \right)}\]
\[ = \frac{1}{\cos \left( 2x + 1 \right)} \lim_{h \to 0} \frac{\sin \left( 2h \right)}{2h} \times 2 \lim_{h \to 0} \frac{1}{\cos \left( 2x + 2h + 1 \right)}\]
\[ = \frac{1}{\cos \left( 2x + 1 \right)} \times 2 \times \frac{1}{\cos \left( 2x + 1 \right)}\]
\[ = \frac{2}{\cos^2 \left( 2x + 1 \right)}\]
\[ = 2 \sec^2 \left( 2x + 1 \right)\]
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