Question

If a function is given as \[y={{\left( {{\tan }^{-1}}x \right)}^{2}}\] , show that \[{{\left( {{x}^{2}}+1 \right)}^{2}}{{y}_{2}}+2x\left( {{x}^{2}}+1 \right){{y}_{1}}=2\] .

Answer 1

Hint: In this question \[{{y}_{1}}\] and \[{{y}_{2}}\] represents first order differentiation and second order differentiation. Differentiate the equation \[y={{\left( {{\tan }^{-1}}x \right)}^{2}}\] with respect to x using formulas \[\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}\] and \[\dfrac{d(ta{{n}^{-1}}x)}{dx}=\dfrac{1}{1+{{x}^{2}}}\] . We know the formula, \[\dfrac{d(uv)}{dx}=v\dfrac{du}{dx}+u\dfrac{dv}{dx}\] .Now, again differentiate the equation \[{{y}_{1}}\left( 1+{{x}^{2}} \right)=2\left( {{\tan }^{-1}}x \right)\] and then solve further.

Complete step-by-step answer:
According to the equation, we have,
\[y={{\left( {{\tan }^{-1}}x \right)}^{2}}\] ……………..(1)
We have to prove, \[{{\left( {{x}^{2}}+1 \right)}^{2}}{{y}_{2}}+2x\left( {{x}^{2}}+1 \right){{y}_{1}}=2\] .
Here, \[{{y}_{1}}\] and \[{{y}_{2}}\] represents first order differentiation and second order differentiation.
That is, \[{{y}_{1}}=\dfrac{dy}{dx}\] ………………………(2)
\[{{y}_{2}}=\dfrac{d{{y}_{2}}}{dx}\] ………………….(3)
Now, using chain rule, differentiating equation (1) with respect to x, we get
\[\begin{align}
  & y={{\left( {{\tan }^{-1}}x \right)}^{2}} \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{d{{\left( {{\tan }^{-1}}x \right)}^{2}}}{dx} \\
\end{align}\]
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{d{{\left( {{\tan }^{-1}}x \right)}^{2}}}{d({{\tan }^{-1}}x)}\times \dfrac{d({{\tan }^{-1}}x)}{dx}\] ………………..(4)
We know the formula, \[\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}\] and \[\dfrac{d(ta{{n}^{-1}}x)}{dx}=\dfrac{1}{1+{{x}^{2}}}\] . Using these two formulas in equation (4), we get
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{d{{\left( {{\tan }^{-1}}x \right)}^{2}}}{d({{\tan }^{-1}}x)}\times \dfrac{d({{\tan }^{-1}}x)}{dx}\]
\[\begin{align}
  & \Rightarrow \dfrac{dy}{dx}=2{{\left( {{\tan }^{-1}}x \right)}^{2-1}}\times \dfrac{1}{1+{{x}^{2}}} \\
 & \Rightarrow \dfrac{dy}{dx}=2\left( {{\tan }^{-1}}x \right)\times \dfrac{1}{1+{{x}^{2}}} \\
\end{align}\]
Using equation (1), we can write the above equation as
\[\Rightarrow {{y}_{1}}\left( 1+{{x}^{2}} \right)=2\left( {{\tan }^{-1}}x \right)\] ………………………….(5)
We know the formula, \[\dfrac{d(uv)}{dx}=v\dfrac{du}{dx}+u\dfrac{dv}{dx}\] .
Using this formula and differentiating equation (5), we get
\[\begin{align}
  & \Rightarrow {{y}_{1}}\left( 1+{{x}^{2}} \right)=2\left( {{\tan }^{-1}}x \right) \\
 & \Rightarrow \dfrac{d{{y}_{1}}}{dx}\left( 1+{{x}^{2}} \right)+{{y}_{1}}\dfrac{d\left( 1+{{x}^{2}} \right)}{dx}=2\dfrac{d\left( {{\tan }^{-1}}x \right)}{dx} \\
\end{align}\]
We know the formula, \[\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}\] , \[\dfrac{d(ta{{n}^{-1}}x)}{dx}=\dfrac{1}{1+{{x}^{2}}}\] and equation (3), we get
\[\begin{align}
  & \Rightarrow {{y}_{2}}\left( 1+{{x}^{2}} \right)+{{y}_{1}}(2x)=2\dfrac{1}{1+{{x}^{2}}} \\
 & \Rightarrow {{y}_{2}}{{\left( 1+{{x}^{2}} \right)}^{2}}+2x\left( 1+{{x}^{2}} \right){{y}_{1}}=2 \\
\end{align}\]
So, LHS = RHS.
Hence, proved.

Note: In this question, one might get confused because there is no any information given for
\[{{y}_{1}}\] and \[{{y}_{2}}\] . Here, \[{{y}_{1}}\] and \[{{y}_{2}}\] are the first order differentiation and second order differentiation respectively. We can also solve this question by just putting the values of \[{{y}_{1}}\] and \[{{y}_{2}}\] in the equation \[{{y}_{2}}{{\left( 1+{{x}^{2}} \right)}^{2}}+2x\left( 1+{{x}^{2}} \right){{y}_{1}}\] .