Logarithmic Differentiation As Seen In Video…

In this video, Patrick JMT demonstrated how to do logarithmic differentiation.

Now I’m going to show you how I’d solve the same problem…

Firstly we must know that: $y=x\ln { \left( \ln { x } \right) } =u\cdot v\\ \\ If\quad y=u\cdot v,\\ \\ \frac { dy }{ dx } =u\frac { dv }{ dx } +v\frac { du }{ dx } \\ \\ y=x\ln { \left( \ln { x } \right) } =u\cdot v\\ \\ u=x,\quad \frac { du }{ dx } =1\\ \\ v=\ln { \left( \ln { x } \right) } =\ln { q } \\ \\ \frac { dv }{ dq } =\frac { 1 }{ q } =\frac { 1 }{ \ln { x } } \\ \\ q=\ln { x } ,\quad \frac { dq }{ dx } =\frac { 1 }{ x } \\ \\ \therefore \quad \frac { dv }{ dx } =\frac { 1 }{ x\ln { x } } \\ \\ \therefore \quad \frac { dy }{ dx } =x\cdot \frac { 1 }{ x\ln { x } } +\ln { \left( \ln { x } \right) } \\ \\ =\frac { 1 }{ \ln { x } } +\ln { \left( \ln { x } \right) }$

Now: $y={ \left( \ln { x } \right) }^{ x }\\ \\ \ln { y=\ln { \left( { \left( \ln { x } \right) }^{ x } \right) } } \\ \\ \ln { y } =x\ln { \left( \ln { x } \right) } \\ \\ \frac { 1 }{ y } \cdot \frac { dy }{ dx } =\frac { 1 }{ \ln { x } } +\ln { \left( \ln { x } \right) } \\ \\ y\cdot \frac { 1 }{ y } \cdot \frac { dy }{ dx } =y\left\{ \frac { 1 }{ \ln { x } } +\ln { \left( \ln { x } \right) } \right\} \\ \\ \frac { dy }{ dx } ={ \left( \ln { x } \right) }^{ x }\left\{ \frac { 1 }{ \ln { x } } +\ln { \left( \ln { x } \right) } \right\}$