It y = {(x+1)²{x-1}}{(x+4)³ e^x} , then dy/dx =

Correct Answer is : (x+1)2x−1(x+4)3ex{(x+1)²{x-1}}{(x+4)³ e^x}(x+4)3ex(x+1)2x−1 [2x+1+12(x−1)]≤ft[ 2/x+1 + 1/2(x-1) ][x+12+2(x−1)1] - [3x+4−1]≤ft[ 3/x+4 - 1 ][x+43−1]

{(x+1)³{x-1}}{(x+4)³ e^x} ≤ft[ 2/x+1 + 1/2(x-1) ] - ≤ft[3/x+4 - 1]

{(x+1)²{x-1}}{(x+4)² e^x} ≤ft[ 2/x+1 + 1/2(x-1) ] - ≤ft[3/x+4 - 1 ]

{(x+1)²{x-1}}{(x+4)³ e^x} ≤ft[ 2/x+1 + 1/2(x-1) ] - ≤ft[ 3/x+4 - 1 ]

{(x+1){x-1}}{(x+4)² e^x} ≤ft[ 2/x+1 + 1/2(x-1) ] - ≤ft[3/x+4 - 1 ]

It y = {(x+1)²{x-1}}{(x+4)³ e^x} , then dy/dx =

Correct Answer is : (x+1)2x−1(x+4)3ex{(x+1)²{x-1}}{(x+4)³ e^x}(x+4)3ex(x+1)2x−1 [2x+1+12(x−1)]≤ft[ 2/x+1 + 1/2(x-1) ][x+12+2(x−1)1] - [3x+4−1]≤ft[ 3/x+4 - 1 ][x+43−1]

Test Index

TS EAMCET 5 May 2018 Shift 1 Solved Paper

Section: Physics

Question : 105 of 160

Marks: +1, -0

It y =

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

, then

\frac{dy}{dx}

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

Validate

Solution:

The given function is,

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

Take log on both side of the above equation.

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

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

- 3 \log (x+4) - x

Differentiate the above equation with respect to x .

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

- \frac{3}{x+4} - 1

\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)

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