If y=frac{(x+1)^2 sqrt{x-1}}{(x+4)^3 e^x}, th | vedclass

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(C) Given $y=\frac{(x+1)^2 \sqrt{x-1}}{(x+4)^3 e^x}$.Taking the natural logarithm on both sides:$\log y = \log \left( \frac{(x+1)^2 (x-1)^{1/2}}{(x+4)

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$\frac{(x+1)^2 \sqrt{x-1}}{(x+4)^3 e^x}\left[\frac{2}{x+1}+\frac{1}{2(x-1)}-\frac{3}{x+4}-1\right]$

Vedclass · Answer

(C) Given $y=\frac{(x+1)^2 \sqrt{x-1}}{(x+4)^3 e^x}$.Taking the natural logarithm on both sides:$\log y = \log \left( \frac{(x+1)^2 (x-1)^{1/2}}{(x+4)^3 e^x} \right)$Using logarithmic properties $\log(ab) = \log a + \log b$ and $\log(a/b) = \log a - \log b$:$\log y = 2 \log(x+1) + \frac{1}{2} \log(x-1) - 3 \log(x+4) - x \log e$Since $\log e = 1$, we have:$\log y = 2 \log(x+1) + \frac{1}{2} \log(x-1) - 3 \log(x+4) - x$Differentiating both sides with respect to $x$:$\frac{1}{y} \frac{dy}{dx} = \frac{2}{x+1} + \frac{1}{2(x-1)} - \frac{3}{x+4} - 1$Therefore, $\frac{dy}{dx} = y \left[ \frac{2}{x+1} + \frac{1}{2(x-1)} - \frac{3}{x+4} - 1 \right]$Substituting the value of $y$:$\frac{dy}{dx} = \frac{(x+1)^2 \sqrt{x-1}}{(x+4)^3 e^x} \left[ \frac{2}{x+1} + \frac{1}{2(x-1)} - \frac{3}{x+4} - 1 \right]$

If $y=\frac{(x+1)^2 \sqrt{x-1}}{(x+4)^3 e^x}$, then $\frac{d y}{d x}=$

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