Question

`(ln2x)^(ln3x)`
how would one differentiate this using logarithmic differentiation?​

Answer 1

`\text{Let,}\\\\~~~~~~~~y = (\ln 2x )^(\ln 3x)\\\\\implies \ln y = \ln\left[(\ln 2x)^(\ln 3x) \right]\\\\\implies \ln y =\ln (3x) \ln( \ln 2x)\\ \\\implies (d)/(dx)( \ln y) = (d)/(dx)\left[ \ln (3x) \ln(\ln 2x) \right]\\\\\implies \frac 1y\cdot(dy)/(dx) =\ln(3x) (d)/(dx) \ln(\ln 2x) + \ln(\ln 2x) (d)/(dx)( \ln 3x)\\\\\implies (dy)/(dx) = \left[\ln(3x) \cdot \frac 1{\ln 2x} \cdot \frac 1{2x} \cdot 2 + \ln(\ln 2x) \cdot \frac 1{3x} \cdot 3\right]y`

`\implies (dy)/(dx) = \left[\ln(3x) \cdot \frac 1{\ln 2x} \cdot \frac 1{2x} \cdot 2 + \ln(\ln 2x) \cdot \frac 1{3x} \cdot 3\right]y\\\\\\\implies (dy)/(dx) = \left[(\ln(3x))/(x \ln(2x))+ (\ln(\ln 2x))/(x)\right](\ln 2x)^(\ln 3x)`