If y=2 x^{n+1}+frac{3}{x^{n}},then x^{2} frac | vedclass

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(B) Given,$y = 2 x^{n+1} + 3 x^{-n} \dots (i)$ Differentiating with respect to $x$: $\frac{dy}{dx} = 2(n+1)x^n + 3(-n)x^{-n-1} = 2(n+1)x^n - 3nx^{-n-1

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$n(n+1) y$

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(B) Given,$y = 2 x^{n+1} + 3 x^{-n} \dots (i)$ Differentiating with respect to $x$: $\frac{dy}{dx} = 2(n+1)x^n + 3(-n)x^{-n-1} = 2(n+1)x^n - 3nx^{-n-1}$ Differentiating again with respect to $x$: $\frac{d^2y}{dx^2} = 2(n+1)(n)x^{n-1} - 3n(-n-1)x^{-n-2}$ $\frac{d^2y}{dx^2} = 2n(n+1)x^{n-1} + 3n(n+1)x^{-n-2}$ $\frac{d^2y}{dx^2} = n(n+1) [2x^{n-1} + 3x^{-n-2}]$ Multiplying both sides by $x^2$: $x^2 \frac{d^2y}{dx^2} = n(n+1) [2x^{n+1} + 3x^{-n}]$ Substituting $y$ from equation $(i)$: $x^2 \frac{d^2y}{dx^2} = n(n+1)y$

If $y=2 x^{n+1}+\frac{3}{x^{n}}$,then $x^{2} \frac{d^{2} y}{d x^{2}}$ is

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