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प्रश्न
tan2 x
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उत्तर
\[\frac{d}{dx}\left( f(x) \right) = \lim_{h \to 0} \frac{f\left( x + h \right) - f\left( x \right)}{h}\]
\[ = \lim_{h \to 0} \frac{\tan^2 \left( x + h \right) - \tan^2 x}{h}\]
\[ = \lim_{h \to 0} \frac{\left[ \tan \left( x + h \right) + \tan x \right]\left[ \tan \left( x + h \right) - \tan x \right]}{h}\]
\[ = \lim_{h \to 0} \frac{\left[ \frac{\sin \left( x + h \right)}{\cos \left( x + h \right)} + \frac{\sin x}{\cos x} \right]\left[ \frac{\sin (x + h)}{\cos (x + h)} - \frac{\sin x}{\cos x} \right]}{h}\]
\[ = \lim_{h \to 0} \frac{\left[ \sin \left( x + h \right) \cos x + \cos \left( x + h \right) \sin x \right]\left[ \sin \left( x + h \right) \cos x - \cos \left( x + h \right) \sin x \right]}{h \cos^2 x \cos^2 \left( x + h \right)}\]
\[ = \lim_{h \to 0} \frac{\left[ \sin \left( 2x + h \right) \right]\left[ \sin h \right]}{h \cos^2 x \cos^2 \left( x + h \right)}\]
\[ = \frac{1}{\cos^2 x} \lim_{h \to 0} \sin \left( 2x + h \right) \lim_{h \to 0} \frac{\sin h}{h} \lim_{h \to 0} \frac{1}{\cos^2 \left( x + h \right)}\]
\[ = \frac{1}{\cos^2 x} \sin \left( 2x \right) \left( 1 \right)\frac{1}{\cos^2 x}\]
\[ = \frac{1}{\cos^2 x} 2 \sin x \cos x \frac{1}{\cos^2 x}\]
\[ = 2 \times \frac{\sin x}{\cos x} \times \frac{1}{\cos^2 x}\]
\[ = 2 \tan x \sec^2 x\]
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