d/dx {(x+1)² {x-1}}{(x+4)³ e^x} = f(x) ≤ft[ 2/x+1 + 1/2(x-1) - 3/x+4 - 1 ] , then f(5)=

Correct Answer is : 881e5{8}{81 e^{5}}81e58

Correct Answer is : 881e5{8}{81 e^{5}}81e58

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AP EAPCET 07-Jul-2022 Shift 1 Paper

Section: Mathematics

Question : 63 of 160

Marks: +1, -0

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

f(5)=

Solution: 👈: Video Solution

\text{Let } y=f(x)=\frac{(x+1)^2 \sqrt{x-1}}{(x+4)^3 e^x}

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

\Rightarrow \log y = 2 \log (x+1) + \frac{1}{2} \log (x-1)

-3 \log (x+4) - x

On differentiating both sides

\text{w.r.t. } x

\frac{1}{y} \frac{dy}{dx} = \frac{2}{x+1} + \frac{1}{2(x-1)} - \frac{3}{x+4} - 1

\Rightarrow \frac{dy}{dx} = f(x) \left[ \frac{2}{x+1} + \frac{1}{2(x-1)} - \frac{3}{x-4} - 1 \right]

\Rightarrow f(x) = \frac{(x+1)^2 \sqrt{x-1}}{(x+4)^3 e^x}

\therefore f(5) = \frac{36 \times 2}{9^3 \cdot e^5} = \frac{8}{81 e^5}

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