The derivative of $f\left(x\right)=\tan (ax+b)$ is

We have, $\begin{array}{l}{f}^{\mathrm{\prime }}\left(x\right)=\underset{h\to 0}{lim} \frac{f(x+h)-f\left(x\right)}{h}\\ =\underset{h\to 0}{lim} \frac{\tan \left[a\right(x+h)+b]-\tan (ax+b)}{h}\\ =\underset{h\to 0}{lim} \frac{\frac{\sin (ax+ah+b)}{\cos (ax+ah+b)}-\frac{\sin (ax+b)}{\cos (ax+b)}}{h}\\ =\underset{h\to 0}{lim} \frac{\sin (ax+ah+b)\cos (ax+b)-\sin (ax+b)}{\cos (ax+ah+b)}\\ =\underset{h\to 0}{lim} \frac{a\cos (ax+b)\cos (ax+ah+b)}{a\cdot h\cos (ax+b)\cos (ax+ah+b)}\\ =\underset{h\to 0}{lim} \frac{a}{\cos (ax+b)\cos (ax+ah+b)}\frac{\mathrm{lax}h\to 0}{a}\frac{\sin ah}{ah}\\ =\frac{{\cos }^2(ax+b)}{h}=a{\sec }^2(ax+b)\end{array}$