Chapter 6 - Inverse Functions - 6.4 Derivatives of Logarithmic Functions - 6.4 Exercises - Page 436: 23

$\sec^2[\ln(ax+b)]*\frac{a}{ax+b}$

$\frac{d}{dx}\tan[\ln(ax+b)]$ $=\sec^2[\ln(ax+b)]*\frac{d}{dx}\ln(ax+b)$ $=\sec^2[\ln(ax+b)]*\frac{1}{ax+b}*\frac{d}{dx}(ax+b)$ $=\sec^2[\ln(ax+b)]*\frac{1}{ax+b}*a$ $=\sec^2[\ln(ax+b)]*\frac{a}{ax+b}$