TS EAMCET · Maths · Differentiation
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 \(\frac{(x+1)^2 \sqrt{x-1}}{(x+4)^3 e^x}\left[\frac{2}{x+1}+\frac{1}{2(x-1)}+\frac{3}{x+4}-1\right]\)
- C \(\frac{(x+1)^2 \sqrt{x-1}}{(x+4)^3 e^x}\left[\frac{2}{x+1}+\frac{1}{2(x-1)}-\frac{3}{x+4}-1\right]\)
- D \(\frac{(x+1) \sqrt{x-1}}{(x+4)^2 e^x}\left[\frac{2}{x+1}+\frac{1}{x-1}-\frac{3}{4+x}-1\right]\)
Answer & Solution
Correct Answer
(C) \(\frac{(x+1)^2 \sqrt{x-1}}{(x+4)^3 e^x}\left[\frac{2}{x+1}+\frac{1}{2(x-1)}-\frac{3}{x+4}-1\right]\)
Step-by-step Solution
Detailed explanation
We have, \[ y=\frac{(x+1)^2 \sqrt{x-1}}{(x+4)^3 e^x} \] Taking log both side, we get…
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