# If  y = (tan^(-1)x)^2  then show that (1+x^2)^2y'' + 2x (1+x^2) y' = 2?

May 20, 2017

To proceed we will need some standard Calculus results:

$\frac{d}{\mathrm{dx}} {\tan}^{- 1} x = \frac{1}{1 + {x}^{2}}$

Now we have:

$y = {\left({\tan}^{- 1} x\right)}^{2}$

If we apply the chain rule then we get:

$y ' = 2 \cdot \left({\tan}^{- 1} x\right) \cdot \frac{1}{1 + {x}^{2}}$
$\setminus \setminus \setminus \setminus = \frac{2 {\tan}^{- 1} x}{1 + {x}^{2}}$

And differentiating again and applying the quotient rule, along with the chain rule, we get:

$y ' ' = \frac{\left(1 + {x}^{2}\right) \left(\frac{d}{\mathrm{dx}} 2 {\tan}^{- 1} x\right) - \left(2 {\tan}^{- 1} x\right) \left(\frac{d}{\mathrm{dx}} \left(1 + {x}^{2}\right)\right)}{1 + {x}^{2}} ^ 2$

$\setminus \setminus \setminus \setminus = \frac{\left(1 + {x}^{2}\right) \left(\frac{2}{1 + {x}^{2}}\right) - \left(2 {\tan}^{- 1} x\right) \left(2 x\right)}{1 + {x}^{2}} ^ 2$

$\setminus \setminus \setminus \setminus = \frac{2 - 2 x \cdot \left(2 {\tan}^{- 1} x\right)}{1 + {x}^{2}} ^ 2$

$\therefore {\left(1 + {x}^{2}\right)}^{2} y ' ' = 2 - 2 x \cdot \left(2 {\tan}^{- 1} x\right)$

$\therefore {\left(1 + {x}^{2}\right)}^{2} y ' ' = 2 - 2 x \left(1 + {x}^{2}\right) \cdot \frac{2 {\tan}^{- 1} x}{1 + {x}^{2}}$

$\therefore {\left(1 + {x}^{2}\right)}^{2} y ' ' = 2 - 2 x \left(1 + {x}^{2}\right) y '$

$\therefore {\left(1 + {x}^{2}\right)}^{2} y ' ' + 2 x \left(1 + {x}^{2}\right) y ' = 2$ QED