Question

="absmiddle" class="latex-formula">
how would one differentiate this using logarithmic differentiation?​

Answer 1

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

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