## Calculus with Applications (10th Edition)

${y^,} = \frac{1}{{2\sqrt {\ln \left| {x - 3} \right|} \,\left( {x - 3} \right)}}$
$\begin{gathered} y = \sqrt {\ln \left| {x - 3} \right|} \hfill \\ \operatorname{Re} write\,\,\,\sqrt {\ln \left| {x - 3} \right|} \,\, = \,{\left( {\ln \left| {x - 3} \right|} \right)^{1/2}} \hfill \\ y = \,\,{\left[ {\ln \left| {x - 3} \right|} \right]^{1/2}} \hfill \\ Use\,\,the\,\,chain\,\,rule \hfill \\ {y^,} = \frac{1}{2}\,\,{\left[ {\ln \left| {x - 3} \right|} \right]^{1/2 - 1}}\,\,{\left[ {\ln \left| {x - 3} \right|} \right]^,} \hfill \\ Use\,\,the\,\,\,formula \hfill \\ \frac{d}{{dx}}\,\,\left[ {\ln g\,\left( x \right)} \right] = \frac{{{g^,}\,\left( x \right)}}{{g\,\left( x \right)}} \hfill \\ Then \hfill \\ {y^,} = \frac{1}{2}\,\,{\left[ {\ln \left| {x - 3} \right|} \right]^{ - 1/2}}\,\left( {\frac{1}{{x - 3}}} \right) \hfill \\ Multiplying \hfill \\ {y^,} = \frac{{\,{{\left( {\ln \left| {x - 3} \right|} \right)}^{ - 1/2}}}}{{2\,\left( {x - 3} \right)}} \hfill \\ {y^,} = \frac{1}{{2\sqrt {\ln \left| {x - 3} \right|} \,\left( {x - 3} \right)}} \hfill \\ \hfill \\ \end{gathered}$