Question

Calculus question! id="TexFormula1" title=" \frac{d}{dx} [(\frac{1}{5x})^{4x}]" alt=" \frac{d}{dx} [(\frac{1}{5x})^{4x}]" align="absmiddle" class="latex-formula">

Answer 1

First, some rewriting:

$\dfrac{\mathrm d}{\mathrm dx}(5x)^{-4x}=\dfrac{\mathrm d}{\mathrm dx}\exp\left(\ln(5x)^{-4x}\right)=\dfrac{\mathrm d}{\mathrm dx}\exp\left(-4x\ln(5x)\right)$

Now taking the derivative is just a matter of applying the chain rule. Since $\dfrac{\mathrm d}{\mathrm dx}e^{f(x)}=\dfrac{\mathrm df(x)}{\mathrm dx}e^{f(x)}$, you end up with

$\dfrac{\mathrm d}{\mathrm dx}(5x)^{-4x}=\dfrac{\mathrm d}{\mathrm dx}[-4x\ln(5x)]\exp\left(-4x\ln(5x)\right)=\dfrac{\mathrm d}{\mathrm dx}[-4x\ln(5x)](5x)^{-4x}$

The product rule tells you that

$\dfrac{\mathrm d}{\mathrm dx}[-4x\ln(5x)]=-4x\dfrac{\mathrm d}{\mathrm dx}[\ln(5x)]+\ln(5x)\dfrac{\mathrm d}{\mathrm dx}[-4x]=-4x\times\dfrac5{5x}-4\ln(5x)=-4(1+\ln(5x))$

So the derivative of the original function is

$\dfrac{\mathrm d}{\mathrm dx}(5x)^{-4x}=-4(1+\ln(5x))(5x)^{-4x}$