Find $y^\prime$ if:

Solution (a) We use the Power Rule for Functions. Then $$\begin{eqnarray*} && y^\prime =\dfrac{d}{dx}( \ln x) ^{2} \underset{\underset{\color{#0066A7}{\hbox{\(\tfrac{d}{dx}[u(x) ] ^{n}{=n}[u(x) ]^{n-1}\tfrac{du}{dx}\)}}}{\color{#0066A7}{\uparrow}}}{=} 2\ln x\cdot \dfrac{d}{dx}\ln x= (2\ln x)\left(\dfrac{1}{x}\right)=\dfrac{2\ln x}{x}\\ \end{eqnarray*}$$

(b) Remember that $\log x=\log _{10}x$. Then $y=\dfrac{1}{2}\log x=\dfrac{1}{2}\log_{10} x$. $$\begin{eqnarray*} && y^\prime =\dfrac{1}{2}\left( \dfrac{d}{dx}\log_{10} x\right) = \dfrac{1}{2}\cdot \dfrac{1}{x\ln 10}=\dfrac{1}{2x\ln 10}\qquad {\color{#0066A7}{\tfrac{d}{dx}\log_{a}{x} = \tfrac{1}{x\ln a}}} \end{eqnarray*}$$

(c) We assume $y$ is a differentiable function of $x$ and use implicit differentiation. Then $$\begin{eqnarray*} \begin{array}{rl@{\quad}l} \dfrac{d}{dx}\left( \ln x+\ln y\right) &= \dfrac{d}{dx}(2x) \\ \dfrac{d}{dx}\ln x+\dfrac{d}{dx}\ln y &= 2 \\ \dfrac{1}{x}+\dfrac{1}{y}\dfrac{dy}{dx} &= 2 & {\color{#0066A7}{\tfrac{d}{dx}\ln {y}=\tfrac{1}{y}\tfrac{dy}{dx}}}\\ y^\prime &= \dfrac{dy}{dx}=y\left(2-\dfrac{1}{x}\right) =\dfrac{y\left(2x-1\right)}{x} & {\color{#0066A7}{\hbox{Solve for}\quad {y^\prime }=\tfrac{dy}{dx}.}} \end{array} \end{eqnarray*}$$