$\dfrac{\mathrm d}{\mathrm dx}\ln x=\dfrac1x\quad(x0)$ Applying the chain rule For \(y=\l

Question

$\dfrac{\mathrm d}{\mathrm dx}\ln x=\dfrac1x\quad(x>0)$ Applying the chain rule 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}$

Accepted Answer

$\frac{\mathrm{d}}{\mathrm{d}x}\ln x=\frac{1}{x}$, $\frac{\mathrm{d}}{\mathrm{d}x}\ln(ax+b)=\frac{a}{ax+b}$

Basics of Logarithmic Derivative

Basics of Logarithmic Derivative

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