Question
Differentiate the following w.r.t. x:
$\frac{8^\text{x}}{\text{x}^8}$

Answer

Let $\text{y}=\frac{8^\text{x}}{\text{x}^8}$
$\Rightarrow\ \log\text{y}=\log\frac{8^\text{x}}{\text{x}^8}$
$\Rightarrow\ \frac{\text{d}}{\text{dy}}\log\text{y}\cdot\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big[\log8^\text{x}-\log\text{x}^8\big]$
$\Rightarrow\ \frac{1}{\text{y}}\cdot\frac{\text{dy}}{\text{dx}}=\big[\text{x}.\log8-8.\log\text{x}]$
On differentiating w.r.t. x, we get
$\Rightarrow\ \frac{1}{\text{y}}\cdot\frac{\text{dy}}{\text{dx}}=\log8.1-8.\frac{1}{\text{x}}$
$\Rightarrow\ \frac{1}{\text{y}}\cdot\frac{\text{dy}}{\text{dx}}=\log8-\frac{8}{\text{x}}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\text{y}\Big(\log8-\frac{8}{\text{x}}\Big)=\frac{8^\text{x}}{\text{x}^8}\Big(\log8-\frac{8}{\text{x}}\Big)$

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