Solve the following

Let $y=\frac{8^{x}}{x^{8}}$

Taking log on both sides, we get $\log y=\log \frac{8^{x}}{x^{8}}$

$\Rightarrow \log y=\log 8^{x}-\log x^{8} \Rightarrow \log y=x \log 8-8 \log x$

Differentiating both sides w.r.t. $x$

$\frac{1}{y} \cdot \frac{d y}{d x}=\log 8.1-\frac{8}{x} \Rightarrow \frac{d y}{d x}=y\left[\log 8-\frac{8}{x}\right]$

Thus, $\frac{d y}{d x}=\frac{8^{x}}{x^{8}}\left[\log 8-\frac{8}{x}\right]$