Q105 / 160 · TS EAMCET 2018 (05 May Shift 1) · Mathematics
If $y=\frac{(x+1)^2 \sqrt{x-1}}{(x+4)^3 e^x}$, then $\frac{d y}{d x}=$
- A$\frac{(x+1)^3 \sqrt{x-1}}{(x+4)^2 e^x}\left[\frac{2}{x+1}+\frac{1}{2(x-1)}-\frac{3}{x+4}-1\right]$
- B$\frac{(x+1)^2 \sqrt{x-1}}{(x+4)^3 e^x}\left[\frac{2}{x+1}+\frac{1}{2(x-1)}+\frac{3}{x+4}-1\right]$
- Correct answerC$\frac{(x+1)^2 \sqrt{x-1}}{(x+4)^3 e^x}\left[\frac{2}{x+1}+\frac{1}{2(x-1)}-\frac{3}{x+4}-1\right]$
- D$\frac{(x+1) \sqrt{x-1}}{(x+4)^2 e^x}\left[\frac{2}{x+1}+\frac{1}{x-1}-\frac{3}{4+x}-1\right]$
Solution
We have,
$$
y=\frac{(x+1)^2 \sqrt{x-1}}{(x+4)^3 e^x}
$$
Taking log both side, we get
$$
\begin{aligned}
& \log y=\log \left(\frac{(x+1)^2 \sqrt{x-1}}{(x+4)^3 \cdot e^x}\right) \\
& \begin{aligned}
\Rightarrow \log y=2 \log (x+1)+\frac{1}{2} \log (x-1) \\
-3 \log (x+4)-x \log e
\end{aligned}
\end{aligned}
$$
$$
\Rightarrow \log y=2 \log (x+1)+\frac{1}{2} \log (x-1)-3 \log (x+4)-x
$$
Differentiate w.r.t. ' $x$ '
$$
\Rightarrow \frac{1}{y} \frac{d y}{d x}=\frac{2}{x+1}+\frac{1}{2(x-1)}-\frac{3}{x+4}-1
$$
$$
\begin{aligned}
\Rightarrow & \frac{d y}{d x}=y\left(\frac{2}{x+1}+\frac{1}{2(x-1)}-\frac{3}{x+4}-1\right) \\
& =\frac{(x+1)^2 \sqrt{x-1}}{(x+4)^3 e^x}\left(\frac{2}{x+1}+\frac{1}{2(x-1)}-\frac{3}{x+4}-1\right)
\end{aligned}
$$
Hence, the correct answer is option: 'C'.