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