If y- tan -1 x 2 and x 2 -1 2 d 2 y d x 2 -2x x 2 -1 dy dx -k- then the value of k is

The correct option is B 2 y=(tan^ 1x)^2 dydx=2tan^ 1x1+x^2 (1+x^2)dydx=2tan^ 1x (1+x^2)d^2ydx^2+2xdydx=211+x^2 (1+x^2)^2d^2ydx^2+ ...

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Higher Order Derivatives

If y= tan -...

Question

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

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Similar questions

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}

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

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} = 2

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