d dxln x= 1x (x>0) Applying the chain rule For y=ln(ax+b), the inner function has derivative a. On its domain ax+b>0, dydx = 1ax+b. a = aax+b

\(\dfrac{\mathrm d}{\mathrm dx}\ln x=\dfrac1x\quad(x>0)\)

For \(y=\ln(ax+b)\), the inner function has derivative \(a\). On its domain \(ax+b>0\),

\(\dfrac{\mathrm{d}y}{\mathrm{d}x}\)\({}=\dfrac1{ax+b}\cdot a\)\({}=\dfrac a{ax+b}\)

Derivatives of Logarithmic Functions