Let y-x2-x-12x-2- Then d2 y-d x2 is

The correct option is B 2 [ 3/(x+1)^4 3/(x+1)^3+4/(x+2)^3 ]y=x^2/(x+1)^2(x+2) y=A/x+1+B/(x+1)^2+C/x+2 ⇒ A= 3, B=1, C=4 ∴ y= 3/x+1+1/(x+1)^2+4/x+2 dy/dx=3/(x+ ...

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Byju's Answer

Standard XII

Mathematics

Algebra of Derivatives

Let y=x2/x+12...

Question

Let $y = \frac{x^{2}}{(x + 1)^{2} (x + 2)}$ . Then $\frac{d^{2} y}{d x^{2}}$ is

2 [\frac{3}{(x + 1)^{4}} - \frac{3}{(x + 1)^{3}} + \frac{4}{(x + 2)^{3}}]

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3 [\frac{2}{(x + 1)^{3}} + \frac{4}{(x + 1)^{2}} - \frac{5}{(x + 2)^{3}}]

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[\frac{6}{(x + 1)^{3}} - \frac{4}{(x + 1)^{2}} + \frac{3}{(x + 1)^{3}}]

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[\frac{7}{(x + 1)^{3}} - \frac{3}{(x + 1)^{2}} + \frac{2}{(x + 1)^{3}}]

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Solution

The correct option is A $2 [\frac{3}{(x + 1)^{4}} - \frac{3}{(x + 1)^{3}} + \frac{4}{(x + 2)^{3}}]$
$y = \frac{x^{2}}{(x + 1)^{2} (x + 2)} y = \frac{A}{x + 1} + \frac{B}{(x + 1)^{2}} + \frac{C}{x + 2}$
$\Rightarrow A = - 3, B = 1, C = 4 ∴ y = \frac{- 3}{x + 1} + \frac{1}{(x + 1)^{2}} + \frac{4}{x + 2} \frac{d y}{d x} = \frac{3}{(x + 1)^{2}} - \frac{2}{(x + 1)^{3}} - \frac{4}{(x + 2)^{2}} \Rightarrow \frac{d^{2} y}{d x^{2}} = - \frac{6}{(x + 1)^{3}} + \frac{6}{(x + 1)^{4}} + \frac{8}{(x + 2)^{3}} \Rightarrow \frac{d^{2} y}{d x^{2}} = 2 [- \frac{3}{(x + 1)^{3}} + \frac{3}{(x + 1)^{4}} + \frac{4}{(x + 2)^{3})}]$

Suggest Corrections

Let y=x2(x+1)2(x+2). Then d2ydx2 is

The correct option is A 2[3(x+1)4−3(x+1)3+4(x+2)3]y=x2(x+1)2(x+2)y=Ax+1+B(x+1)2+Cx+2 ⇒A=−3, B=1, C=4∴y=−3x+1+1(x+1)2+4x+2dydx=3(x+1)2−2(x+1)3−4(x+2)2⇒d2ydx2=−6(x+1)3+6(x+1)4+8(x+2)3⇒d2ydx2=2[−3(x+1)3+3(x+1)4+4(x+2)3)]

Let $y = \frac{x^{2}}{(x + 1)^{2} (x + 2)}$ . Then $\frac{d^{2} y}{d x^{2}}$ is