Differentiate the function $\frac{8^x}{x^8}$ with respect to $x$.

$\frac{8^x}{x^8} \left( \log 8 - \frac{8}{x} \right)$

The correct answer is Option (1) → $\frac{8^x}{x^8} \left( \log 8 - \frac{8}{x} \right)$ ##

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

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

On differentiating w.r.t. $x$, we get

$\frac{d}{dy} \log y \cdot \frac{dy}{dx} = \frac{d}{dx} [\log 8^x - \log x^8] \quad \left[ ∵\log \frac{m}{n} = \log m - \log n \right]$

$\Rightarrow \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} [x \cdot \log 8 - 8 \cdot \log x] \quad [∵\log m^n = n \log m]$

$\Rightarrow \frac{1}{y} \frac{dy}{dx} = \log 8 \cdot 1 - 8 \cdot \frac{1}{x} \quad \left[ ∵ \frac{d}{dx} \log x = \frac{1}{x} \right]$

$\Rightarrow \frac{1}{y} \frac{dy}{dx} = \log 8 - \frac{8}{x}$

$∴\frac{dy}{dx} = y \left( \log 8 - \frac{8}{x} \right) = \frac{8^x}{x^8} \left( \log 8 - \frac{8}{x} \right) \quad \left[ ∵y = \frac{8^x}{x^8} \right]$