Respuesta :

[tex]y=\dfrac{8x^4}{\ln x}[/tex]

By the quotient rule, the derivative is

[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\ln x\dfrac{\mathrm d}{\mathrm dx}[8x^4]-8x^4\dfrac{\mathrm d}{\mathrm dx}[\ln x]}{(\ln x)^2}[/tex]
[tex]y'=\dfrac{48x^3\ln x-\dfrac{8x^4}x}{\ln^2x}[/tex]
[tex]y'=\dfrac{48x^3\ln x-8x^3}{\ln^2x}[/tex]
[tex]y'=\dfrac{8x^3(6\ln x-1)}{\ln^2x}[/tex]