If $y={{\left( {{\tan }^{-1}}x \right)}^{2}}$ , show that ${{\left( {{x}^{2}}+1 \right)}^{2}}\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+2x\left( {{x}^{2}}+1 \right)\dfrac{dy}{dx}=2$ .

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Hint: In order to solve this problem, we need to know the standard formulas of differentiation. The formulas are $\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}$ and $\dfrac{d}{dx}\left( {{\tan }^{-1}}x \right)=\dfrac{1}{1+{{x}^{2}}}$ and $\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}$ . We need to find the first order and the second-order differentiation of the given function and substitute in the left-hand side expression to prove that right-hand side expression.

Complete step by step answer:
We are given the expression of $y={{\left( {{\tan }^{-1}}x \right)}^{2}}$ , and we need to prove that ${{\left( {{x}^{2}}+1 \right)}^{2}}\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+2x\left( {{x}^{2}}+1 \right)\dfrac{dy}{dx}=2$ .
Let's differentiate the equation $y={{\left( {{\tan }^{-1}}x \right)}^{2}}$ with respect to x
We need to use the standard formulas such as $\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}$ and $\dfrac{d}{dx}\left( {{\tan }^{-1}}x \right)=\dfrac{1}{1+{{x}^{2}}}$ .
Performing the operation we get,
  & \dfrac{dy}{dx}=\dfrac{d}{dx}{{\left( {{\tan }^{-1}}x \right)}^{2}} \\
 & =2\left( {{\tan }^{-1}}x \right)\dfrac{d}{dx}\left( {{\tan }^{-1}}x \right) \\
 & =2\left( {{\tan }^{-1}}x \right)\left( \dfrac{1}{1+{{x}^{2}}} \right)
As we can see that in the expression that we have to prove we are given the second-order differential equation,
 So, we need to differentiate the expression again.
The standard formula required are as follows,
$\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}$ .
By performing the differentiation with respect to x we get,
  & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( 2\left( {{\tan }^{-1}}x \right)\left( \dfrac{1}{1+{{x}^{2}}} \right) \right) \\
 & =2\left( \dfrac{-{{\tan }^{-1}}x\left( 2x \right)+\left( 1+{{x}^{2}} \right)\left( \dfrac{1}{1+{{x}^{2}}} \right)}{{{\left( 1+{{x}^{2}} \right)}^{2}}} \right) \\
Solving this we get,
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=2\left( \dfrac{-2x\left( {{\tan }^{-1}}x \right)+1}{{{\left( 1+{{x}^{2}} \right)}^{2}}} \right)$
Now, substituting all the values from the expression we get,
  & {{\left( {{x}^{2}}+1 \right)}^{2}}\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+2x\left( {{x}^{2}}+1 \right)\dfrac{dy}{dx} \\
 & {{\left( {{x}^{2}}+1 \right)}^{2}}\left( 2\left( \dfrac{-2x\left( {{\tan }^{-1}}x \right)+1}{{{\left( 1+{{x}^{2}} \right)}^{2}}} \right) \right)+2x\left( {{x}^{2}}+1 \right)\left( 2\left( {{\tan }^{-1}}x \right)\left( \dfrac{1}{1+{{x}^{2}}} \right) \right) \\
Simplifying the equation we get,
  & {{\left( {{x}^{2}}+1 \right)}^{2}}\left( 2\left( \dfrac{-2x\left( {{\tan }^{-1}}x \right)+1}{{{\left( 1+{{x}^{2}} \right)}^{2}}} \right) \right)+2x\left( {{x}^{2}}+1 \right)\left( 2\left( {{\tan }^{-1}}x \right)\left( \dfrac{1}{1+{{x}^{2}}} \right) \right) \\
 & \left( 2\left( \dfrac{-2x\left( {{\tan }^{-1}}x \right)+1}{1} \right) \right)+2x\left( 2\left( {{\tan }^{-1}}x \right) \right) \\
 & -4x{{\tan }^{-1}}x+2+4x{{\tan }^{-1}}x \\
 & =2 \\
Hence left-hand side = right-hand side.
Therefore, we can see that the ${{\left( {{x}^{2}}+1 \right)}^{2}}\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+2x\left( {{x}^{2}}+1 \right)\dfrac{dy}{dx}=2$ is proved

The calculation in this problem is quite complicated. While differentiating we need to use the chain rule. The chain rule is used when we need to find the differentiation of the function inside the function. And it is carried out as we need to differentiate the outermost function followed by the multiplication of the differentiation of the innermost function.
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