If [tex]y=x^{\log_2(x)}[/tex], then taking the logarithm of both sides gives
[tex]\ln(y) = \ln\left(x^{\log_2(x)}\right) = \log_2(x) \ln(x)[/tex]
Differentiate both sides with respect to x :
[tex]\dfrac{d\ln(y)}{dx} = \dfrac{d\log_2(x)}{dx}\ln(x) + \log_2(x)\dfrac{d\ln(x)}{dx}[/tex]
[tex]\dfrac1y \dfrac{dy}{dx} = \dfrac{d\log_2(x)}{dx}\ln(x) + \dfrac{\log_2(x)}{x}[/tex]
Now if [tex]z=\log_2(x)[/tex], then [tex]2^z=x[/tex]. Rewrite
[tex]2^z = e^{\ln(2^z)} = e^{\ln(2)z}[/tex]
Then by the chain rule,
[tex]\dfrac{d2^z}{dx} = \dfrac{dx}{dx}[/tex]
[tex]\dfrac{de^{\ln(2)z}}{dx} = 1[/tex]
[tex]e^{\ln(2)z} \ln(2) \dfrac{dz}{dx}= 1[/tex]
[tex]\dfrac{dz}{dx} = \dfrac{1}{e^{\ln(2)z}\ln(2)}[/tex]
[tex]\dfrac{dz}{dx} = \dfrac{1}{2^z \ln(2)}[/tex]
[tex]\dfrac{d\log_2(x)}{dx} = \dfrac{1}{\ln(2)x}[/tex]
So we have
[tex]\dfrac1y \dfrac{dy}{dx} = \dfrac{\ln(x)}{\ln(2)x}+ \dfrac{\log_2(x)}{x}[/tex]
[tex]\dfrac1y \dfrac{dy}{dx} = \dfrac{\log_2(x)}{x}+ \dfrac{\log_2(x)}{x}[/tex]
[tex]\dfrac1y \dfrac{dy}{dx} = \dfrac{2\log_2(x)}{x}[/tex]
[tex]\dfrac1y \dfrac{dy}{dx} = \dfrac{\log_2(x^2)}{x}[/tex]
[tex]\dfrac{dy}{dx} = \dfrac{y\log_2(x^2)}{x}[/tex]
Replace y :
[tex]\dfrac{dy}{dx} = \dfrac{x^{\log_2(x)}\log_2(x^2)}{x}[/tex]
[tex]\dfrac{dy}{dx} = x^{\log_2(x)-1}\log_2(x^2)[/tex]
