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

& y={{\left( {{\tan }^{-1}}x \right)}^{2}} \\\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{d{{\left( {{\tan }^{-1}}x \right)}^{2}}}{dx} \\\ \end{aligned}$$ $$\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{aligned} & \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{aligned}$$ Using equation (1), we can write the above equation as $$\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{aligned} & \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{aligned}$$ We know the formula, $$\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{aligned} & \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{aligned}$$ So, LHS = RHS. Hence, proved. Note: In this question, one might get confused because there is no any information given for $${{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}}$$ .