\(y=\frac{\sqrt[3]{1+3 x} \sqrt[4]{1+4 x} \sqrt[5]{1+5 x}}{\sqrt[7]{1+7 x} \sqrt[8]{1+8 x}}\). Then \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) at \(x=0\) is

\(y=\frac{\sqrt[3]{1+3 x} \sqrt[4]{1+4 x} \sqrt[5]{1+5 x}}{\sqrt[7]{1+7 x} \sqrt[8]{1+8 x}}\)
\(\therefore \quad \log y=\frac{1}{3} \log (1+3 x)+\frac{1}{4} \log (1+4 x)\)
\(\begin{aligned}+\frac{1}{5} \log (1+5 x) & -\frac{1}{7} \log (1+7 x) \\ & -\frac{1}{8} \log (1+8 x)\end{aligned}\)
Differentiating both sides w.r.t. \(x\), we get
\(
\begin{array}{r}
\frac{1}{y} \cdot \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1}{3} \cdot \frac{1}{1+3 x} \cdot 3+\frac{1}{4} \cdot \frac{1}{1+4 x} \cdot 4+\frac{1}{5} \cdot \frac{1}{1+5 x} \cdot 5 \\
-\frac{1}{7} \cdot \frac{1}{1+7 x} \cdot 7-\frac{1}{8} \cdot \frac{1}{1+8 x} \cdot 8
\end{array}
\)
\(\therefore \frac{1}{y} \cdot \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1}{1+3 x}+\frac{1}{1+4 x}+\frac{1}{1+5 x}-\) \(\frac{1}{1+7 x}-\frac{1}{1+8 x} \)
\( \therefore \frac{1}{1} \cdot\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{x=0}=\frac{1}{1+0}+\frac{1}{1+0}+\frac{1}{1+0}\) \(-\frac{1}{1+0}-\frac{1}{1+0} \)
\( \ldots[\text { At } x=0, y=1]\)
\(\Rightarrow\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{x=0}=1\)