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

Correct Answer is : (x+1)2√x−1(x+4)3ex [2x+1+12(x−1) - 3x+4−1]

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

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

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

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

TS EAMCET 5 May 2018 Shift 1 Solved Paper

Section: Physics

Question : 105 of 160

Marks: +1, -0

It y =

(x+1)2√x−1

(x+4)3ex

, then

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