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Higher Order Derivatives

If y= tan -...

Question

If $y = {({tan}^{- 1} x)}^{2}$ , show that ${(x^{2} + 1)}^{2} y_{2} + 2 x (x^{2} + 1) y_{1} = 2$ .

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Solution

Let

$y = (t a n^{- 1} x)^{2}$

Differentiate w,r.t. $x$ .

$\frac{d y}{d x} = 2. t a n^{- 1} x \frac{1}{(1 + x^{2})}$

$y_{1} (1 + x^{2}) = 2 t a n^{- 1} x$

Again differentiating w.r.t $x$ .

$y_{2} (1 + x^{2}) + y_{1} (2 x) = \frac{2}{(1 + x^{2})}$

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

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Similar questions

y = (t a n^{- 1} x)^{2}

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

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

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

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

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

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

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

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