If y -tan -1 x 2- then the numerical value of x 2-12 d 2 y - dx 2-2 x x 2-1 dy - dx is

Y = (tan^ 1 x)^2 ⇒ y' =2tan^ 1x ×11+x^2 ⇒ (1+x^2)y' = 2tan^ 1 x ⇒ nbsp;(1+x^2)y”+2xy' = 21+x^2 ⇒ (x^2 +1)^2y”+2x(x^2 +1)y' = 2 ...

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Chain Rule of Differentiation

If y =tan -1 ...

Question

If $y = ({tan}^{- 1} x)^{2}$ , then the numerical value of $(x^{2} + 1)^{2} \frac{d^{2} y}{d x^{2}} + 2 x (x^{2} + 1) \frac{d y}{d x}$ is

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y = {({tan}^{- 1} x)}^{2}

{(x^{2} + 1)}^{2} \frac{d^{2} y}{d x^{2}} + 2 x (x^{2} + 1) \frac{d y}{d x} = 2

y = {({tan}^{- 1} x)}^{2}

{(x^{2} + 1)}^{2} \frac{d^{2} y}{d x^{2}} + 2 x (x^{2} + 1) \frac{d y}{d x} =

y = {({tan}^{- 1} x)}^{2}

{(x^{2} + 1)}^{2} \frac{d^{2} y}{d x^{2}} + 2 x (x^{2} + 1) \frac{d y}{d x} = k

k

x \sqrt{1 + y} + y \sqrt{1 + x} = 0

(x + 1)^{2} \frac{d^{2} y}{d x^{2}} + 2 (x + 1) \frac{d y}{d x}

y = 3

y = ({tan}^{- 1} x)^{2}

(x^{2} + 1)^{2} y_{2} + 2 x (x^{2} + 1) y_{1}

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