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