Question

\[\tan \sqrt{x}\] 

Answer 1

\[text{ Let } f(x) = \tan x^2 \]
\[\text{ Thus, we have }: \]
\[f(x + h) = \tan (x + h )^2 \]
\[\frac{d}{dx}\left( f(x) \right) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}\]
\[ = \lim_{h \to 0} \frac{\tan (x + h )^2 - \tan x^2}{h}\]
\[ = \lim_{h \to 0} \frac{\sin\left( \left( x + h \right)^2 - x^2 \right)}{h \cos \left( x + h \right)^2 \cos x^2} \left[ \because \tan A - \tan B = \frac{\sin (A - B)}{\cos A \cos B} \right]\]
\[ = \lim_{h \to 0} \frac{\sin( x^2 + h^2 + 2hx - x^2 )}{h\cos \left( x + h \right)^2 \cos x^2}\]
\[ = \lim_{h \to 0} \frac{\sin(h\left( h + 2x) \right)}{h\left( h + 2x \right) \cos \left( x + h \right)^2 \cos x^2} \times \left( h + 2x \right)\]
\[ = \lim_{h \to 0} \frac{\sin(h\left( h + 2x) \right)}{(h\left( h + 2x) \right)} \lim_{h \to 0} \frac{h + 2x}{\cos(x + h )^2 \cos x^2} \left[ As \lim_{h \to 0} \frac{\sin(h\left( h + 2x) \right)}{(h\left( h + 2x) \right)} = 1 \right]\]
\[ = 1 \times \frac{2x}{\cos^2 x^2}\]
\[ = 2x \sec^2 x^2 \]
\[\]