The derivative of $f (x) = \tan (a x + b)$ is

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$\sec^{2} (a x + b)$

$b \sec^{2} (a x + b)$

$a b \sec^{2} (a x + b)$

$a \sec^{2} (a x + b)$

answer is C.

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Detailed Solution

We have, $\begin{array}{l} f^{'} (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \\ = lim_{h \to 0} \frac{\tan [a (x + h) + b] - \tan (a x + b)}{h} \\ = lim_{h \to 0} \frac{\frac{\sin (a x + a h + b)}{\cos (a x + a h + b)} - \frac{\sin (a x + b)}{\cos (a x + b)}}{h} \\ = lim_{h \to 0} \frac{\sin (a x + a h + b) \cos (a x + b) - \sin (a x + b)}{\cos (a x + a h + b)} \\ = lim_{h \to 0} \frac{a \cos (a x + b) \cos (a x + a h + b)}{a \cdot h \cos (a x + b) \cos (a x + a h + b)} \\ = lim_{h \to 0} \frac{a}{\cos (a x + b) \cos (a x + a h + b)} \frac{lax h \to 0}{a} \frac{\sin a h}{a h} \\ = \frac{\cos^{2} (a x + b)}{h} = a \sec^{2} (a x + b) \end{array}$