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

If $y=\left[\mathrm{e}^{4 x}\left(\frac{x-4}{x+3}\right)^{\frac{3}{4}}\right]$ then $\frac{\mathrm{d} y}{\mathrm{~d} x}=$
  1. $\frac{\mathrm{d} y}{\mathrm{~d} x}=y\left[4+\frac{21}{4(x-4)(x+3)}\right]$
  2. $\frac{\mathrm{d} y}{\mathrm{~d} x}=\left[4+\frac{21}{4(x-4)(x+3)}\right]$
  3. $\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1}{y}\left[4+\frac{21}{4(x-4)(x+3)}\right]$
  4. $\frac{\mathrm{d} y}{\mathrm{~d} x}=y\left[4+\frac{21}{4(x+4)(x+3)}\right]$

Solution

$y=e^{4 x}\left(\frac{x-4}{x+3}\right)^{\frac{3}{4}}$
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, $\begin{aligned} & \frac{1}{y} \frac{\mathrm{~d} y}{\mathrm{~d} x}=4+\frac{3}{4} \times \frac{1}{(x-4)} \cdot \frac{\mathrm{d}}{\mathrm{~d} x}\left(\frac{x-4}{x+3}\right) (x+3) \\ & \frac{1}{y} \frac{\mathrm{~d} y}{\mathrm{~d} x}=4+\frac{3}{4}\left(\frac{x+3}{x-4}\right) \times\left(\frac{(x+3)-(x-4)}{(x+3)^2}\right) \\ & \frac{1}{y} \frac{\mathrm{~d} y}{\mathrm{~d} x}=4+\frac{3}{4}\left(\frac{1}{x-4}\right) \times\left(\frac{x+3-x+4}{(x+3)}\right) \\ & \frac{1}{y} \frac{\mathrm{~d} y}{\mathrm{~d} x}=4+\frac{3}{4} \times\left(\frac{1}{x-4}\right) \times\left(\frac{7}{x+3}\right) \end{aligned}$ $\therefore \quad \frac{\mathrm{d} y}{\mathrm{~d} x}=y\left[4+\frac{21}{4(x-4)(x+3)}\right]$

Asked in: MHT CET 2024 (02 May Shift 1)

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