If \(y=\left[\mathrm{e}^{4 x}\left(\frac{x-4}{x+3}\right)^{\frac{3}{4}}\right]\) then \(\frac{\mathrm{d} y}{\mathrm{~d} x}=\)

\(y=e^{4 x}\left(\frac{x-4}{x+3}\right)^{\frac{3}{4}}\)
Taking log on both sides,
\(\begin{aligned}
& \log y=\log \left[\mathrm{e}^{4 x}\left(\frac{x-4}{x+3}\right)^{\frac{3}{4}}\right] \\
& \log y=\log \mathrm{e}^{4 x}+\log \left(\frac{x-4}{x+3}\right)^{\frac{3}{4}} \\
& \log y=4 x \log \mathrm{e}+\frac{3}{4} \log \left(\frac{x-4}{x+3}\right) \\
& \log y=4 x+\frac{3}{4} \log \left(\frac{x-4}{x+3}\right)
\end{aligned}\)
Differentiating w.r.to \(x\) on both sides,
\(\begin{aligned}
& \frac{1}{y} \frac{\mathrm{~d} y}{\mathrm{~d} x}=4+\frac{3}{4} \times \frac{1}{(x-4)} \cdot \frac{\mathrm{d}}{\mathrm{~d} x}\left(\frac{x-4}{x+3}\right) (x+3) \\
& \frac{1}{y} \frac{\mathrm{~d} y}{\mathrm{~d} x}=4+\frac{3}{4}\left(\frac{x+3}{x-4}\right) \times\left(\frac{(x+3)-(x-4)}{(x+3)^2}\right) \\
& \frac{1}{y} \frac{\mathrm{~d} y}{\mathrm{~d} x}=4+\frac{3}{4}\left(\frac{1}{x-4}\right) \times\left(\frac{x+3-x+4}{(x+3)}\right) \\
& \frac{1}{y} \frac{\mathrm{~d} y}{\mathrm{~d} x}=4+\frac{3}{4} \times\left(\frac{1}{x-4}\right) \times\left(\frac{7}{x+3}\right)
\end{aligned}\)
\(\therefore \quad \frac{\mathrm{d} y}{\mathrm{~d} x}=y\left[4+\frac{21}{4(x-4)(x+3)}\right]\)