Continuity and Differentiability — Class 12 Maths Solution

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

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

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

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

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

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

therefore,$\frac{{dy}}{{dx}} = y\left( {\log 8 - \frac{8}{x}} \right) = \frac{{{8^x}}}{{{x^8}}}\left( {\log 8 - \frac{8}{x}} \right)$