First, some rewriting:
[tex]\dfrac{\mathrm d}{\mathrm dx}(5x)^{-4x}=\dfrac{\mathrm d}{\mathrm dx}\exp\left(\ln(5x)^{-4x}\right)=\dfrac{\mathrm d}{\mathrm dx}\exp\left(-4x\ln(5x)\right)[/tex]
Now taking the derivative is just a matter of applying the chain rule. Since [tex]\dfrac{\mathrm d}{\mathrm dx}e^{f(x)}=\dfrac{\mathrm df(x)}{\mathrm dx}e^{f(x)}[/tex], you end up with
[tex]\dfrac{\mathrm d}{\mathrm dx}(5x)^{-4x}=\dfrac{\mathrm d}{\mathrm dx}[-4x\ln(5x)]\exp\left(-4x\ln(5x)\right)=\dfrac{\mathrm d}{\mathrm dx}[-4x\ln(5x)](5x)^{-4x}[/tex]
The product rule tells you that
[tex]\dfrac{\mathrm d}{\mathrm dx}[-4x\ln(5x)]=-4x\dfrac{\mathrm d}{\mathrm dx}[\ln(5x)]+\ln(5x)\dfrac{\mathrm d}{\mathrm dx}[-4x]=-4x\times\dfrac5{5x}-4\ln(5x)=-4(1+\ln(5x))[/tex]
So the derivative of the original function is
[tex]\dfrac{\mathrm d}{\mathrm dx}(5x)^{-4x}=-4(1+\ln(5x))(5x)^{-4x}[/tex]
