Find \(y^\prime \) if:
(a) \[ \begin{eqnarray*} && y^\prime = \dfrac{d}{dx}\ln (5x) = \dfrac{\dfrac{d}{dx}(5x) }{5x}=\dfrac{5}{5x} = \dfrac{1}{x} \qquad {\color{#0066A7}{\tfrac{d}{dx}\ln u({x}) = \tfrac{u^\prime (x)}{u(x)}}} \end{eqnarray*} \]
(b) \[ \begin{eqnarray*} y^\prime = \dfrac{d}{dx}[ x\ln (x^{2}+1)] &=& x\dfrac{d}{dx} \ln (x^{2}+1) +\left( \dfrac{d}{dx}x\right) \ln (x^{2}+1)\qquad{\color{#0066A7}{\hbox{Product Rule}}} \\ &&\underset{\underset{\color{#0066A7}{\tfrac{d}{dx}\ln u(x)=\tfrac{u^\prime (x)}{u(x)}}}{\color{#0066A7}{\uparrow}}}{=} x\cdot \dfrac{\dfrac{d}{dx}(x^{2}+1) }{ x^{2}+1}+1\cdot \ln (x^{2}+1) =\dfrac{2x^{2}}{x^{2}+1}+\ln (x^{2}+1) \end{eqnarray*} \]
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(c) \[ \begin{eqnarray*} y &=&\ln \vert x\vert =\left\{ \begin{array}{c@{\quad}l} \ln x & \hbox{if}\; x>0 \\ \ln (-x) & \hbox{if}\; x<0 \end{array} \right. \\ y^\prime &=& \dfrac{d}{dx}\ln \vert x\vert =\left\{ \begin{array}{c@{\quad}l} \dfrac{1}{x} & \hbox{if}\;x>0 \\ \dfrac{1}{-x}(-1) =\dfrac{1}{x} & \hbox{if}\;x<0 \end{array}\right. \end{eqnarray*} \]
That is, \(\dfrac{d}{dx}\ln \vert x\vert =\dfrac{1}{x}\) for all numbers \(x≠ 0\).