If \(y=(x-1)^{2}(x-2)^{3}(x-3)^{5}\), then \(\frac{d y}{d x}\) at \(x=4\) is equal to

\(y=(x-1)^{2}(x-2)^{3}(x-3)^{5}\)
Taking log on both sides,
\(\begin{aligned}
&\Rightarrow \log y=\log \left[(x-1)^{2}(x-2)^{3}(x-3)^{5}\right] \\
&\log y=2 \log (x-1)+3 \log (x-2)+5 \log (x-3)
\end{aligned}\)
On both sides differentiating w.r.t. \(x\), we get
\(\frac{1}{y} \frac{d y}{d x}=\frac{2}{x-1}+\frac{3}{x-2}+\frac{5}{x-3}\)
\(\Rightarrow \quad \frac{d y}{d x}=(x-1)^{2}(x-2)^{3}(x-3)^{5}\)
\(\left[\frac{2}{(x-1)}+\frac{3}{(x-2)}+\frac{5}{(x-3)}\right]\)
\(\therefore\left(\frac{d y}{d x}\right)_{x=4}=3^{2} \times 2^{3} \times 1^{5}\left[\frac{2}{3}+\frac{3}{2}+5\right]\)
\(=9 \times 8 \times\left(\frac{4+9+30}{6}\right)=516\)