Differentiation of Logarithmic Functions

Example 1:

$$ y=ln \,3x,\;\;\; \textup{Find} \;\;\frac{dy}{dx}$$

Solution:

\begin{align*} y&=ln \,3x\\ \frac{dy}{dx}&=\frac{d}{dx} ln \,3x\\\\ \textup{Let}\; u&= 3x\\\\ \frac{dy}{dx}&=\frac{d}{dx} ln\,u\\ \frac{dy}{dx}&=\frac{d}{du} ln\,u \frac{du}{dx}\\ \frac{dy}{dx}&=\frac{1}u \frac{d}{dx} (3x)\\ \frac{dy}{dx}&=\frac{1}{3x}(3)\\ \frac{dy}{dx}&=\frac{1}x\\ \end{align*}

Example 2:

$$ y=ln(x+1),\;\;\; \textup{Find} \;\;\frac{dy}{dx}$$

Solution:

\begin{align*} y&=ln(x+1)\\ \frac{dy}{dx}&=\frac{d}{dx} ln (x+1)\\\\ \textup{Let}\; u&= x+1\\\\ \frac{dy}{dx}&=\frac{d}{dx} ln\,u\\ \frac{dy}{dx}&=\frac{d}{du} ln\,u \frac{du}{dx}\\ \frac{dy}{dx}&=\frac{1}u \frac{d}{dx} (x+1)\\ \frac{dy}{dx}&=\frac{1}{x+1}(1)\\ \frac{dy}{dx}&=\frac{1}{x+1}\\ \end{align*}

Example 3:

$$ y=ln(x^2+1),\;\;\; \textup{Find} \;\;\frac{dy}{dx}$$

Solution:

\begin{align*} y&=ln(x^2+1)\\ \frac{dy}{dx}&=\frac{d}{dx} ln (x^2+1)\\\\ \textup{Let}\; u&= x^2+1\\\\ \frac{dy}{dx}&=\frac{d}{dx} ln\,u\\ \frac{dy}{dx}&=\frac{d}{du} ln\,u \frac{du}{dx}\\ \frac{dy}{dx}&=\frac{1}u \frac{d}{dx} (x^2+1)\\ \frac{dy}{dx}&=\frac{1}{x^2+1}(2x)\\ \frac{dy}{dx}&=\frac{2x}{x^2+1}\\ \end{align*}

Example 4:

$$ y=ln(ax+b),\;\;\; \textup{Find} \;\;\frac{dy}{dx}$$

Solution:

\begin{align*} y&=ln(ax+b)\\ \frac{dy}{dx}&=\frac{d}{dx} ln (ax+b)\\\\ \textup{Let}\; u&= ax+b\\\\ \frac{dy}{dx}&=\frac{d}{dx} ln\,u\\ \frac{dy}{dx}&=\frac{d}{du} ln\,u \frac{du}{dx}\\ \frac{dy}{dx}&=\frac{1}u \frac{d}{dx} (ax+b)\\ \frac{dy}{dx}&=\frac{1}{ax+b}\left(\frac{d}{dx}ax+\frac{d}{dx}b\right)\\ \frac{dy}{dx}&=\frac{1}{ax+b}\left(a\frac{d}{dx}x+\frac{d}{dx}b\right)\\\\ \because\frac{d}{dx} x&=1 \;\;\; \textup{and} \;\;\;\frac{d}{dx}c=0\\\\ \frac{dy}{dx}&=\frac{1}{ax+b}(a(1)+0)\\ \frac{dy}{dx}&=\frac{1}{ax+b}(a)\\ \frac{dy}{dx}&=\frac{a}{ax+b}\\ \end{align*}