# If y = tanh^(-1)((2x)/(1+x^2)) then show that the second derivative is (4x)/(1 -x^2)^2?

##### 2 Answers
Jan 31, 2018

We have:

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

From which we have:

$\tanh y = \frac{2 x}{1 + {x}^{2}}$

Differentiating Implicitly, and applying the chain rule, we get:

${\sech}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(1 + {x}^{2}\right) \left(2\right) - \left(2 x\right) \left(2 x\right)}{1 + {x}^{2}} ^ 2$

$\therefore \left(1 - {\tanh}^{2} y\right) \setminus \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 + 2 {x}^{2} - 4 {x}^{2}}{1 + {x}^{2}} ^ 2$

$\therefore \left(1 - {\left(\frac{2 x}{1 + {x}^{2}}\right)}^{2}\right) \setminus \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 - 2 {x}^{2}}{1 + {x}^{2}} ^ 2$

$\therefore \left(\frac{{\left(1 + {x}^{2}\right)}^{2} - 4 {x}^{2}}{1 + {x}^{2}} ^ 2\right) \setminus \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 - 2 {x}^{2}}{1 + {x}^{2}} ^ 2$

$\therefore \left(1 + 2 {x}^{2} + {x}^{4} - 4 {x}^{2}\right) \setminus \frac{\mathrm{dy}}{\mathrm{dx}} = 2 - 2 {x}^{2}$

$\therefore \left(1 - 2 {x}^{2} + {x}^{4}\right) \setminus \frac{\mathrm{dy}}{\mathrm{dx}} = 2 - 2 {x}^{2}$

$\therefore {\left(1 - {x}^{2}\right)}^{2} \setminus \frac{\mathrm{dy}}{\mathrm{dx}} = 2 \left(1 - {x}^{2}\right)$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{1 - {x}^{2}}$

And differentiating again, and applying the chain rule, we gte

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 2 \left(- 1\right) {\left(1 - {x}^{2}\right)}^{- 2} \left(- 2 x\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{4 x}{1 - {x}^{2}} ^ 2 \setminus \setminus$ QED

Feb 1, 2018

We have:

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

Using the standard result:

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

In conjunction with the chain rule and quotient rule, then we get:

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

$\setminus \setminus \setminus \setminus \setminus \setminus = \frac{1}{\frac{{\left(1 + x\right)}^{2} - 4 {x}^{2}}{{\left(1 + {x}^{2}\right)}^{2}}} \left\{\frac{\left(1 + {x}^{2}\right) \left(2\right) - \left(2 x\right) \left(2 x\right)}{1 + {x}^{2}} ^ 2\right\}$

$\setminus \setminus \setminus \setminus \setminus \setminus = \frac{{\left(1 + {x}^{2}\right)}^{2}}{\left({\left(1 + {x}^{2}\right)}^{2} - 4 {x}^{2}\right)} \left\{\frac{\left(1 + {x}^{2}\right) \left(2\right) - \left(2 x\right) \left(2 x\right)}{1 + {x}^{2}} ^ 2\right\}$

$\setminus \setminus \setminus \setminus \setminus \setminus = 2 \setminus \frac{1 + {x}^{2} - 2 {x}^{2}}{\left(1 + 2 {x}^{2} + {x}^{4} - 4 {x}^{2}\right)}$

$\setminus \setminus \setminus \setminus \setminus \setminus = 2 \setminus \frac{1 - {x}^{2}}{1 - 2 {x}^{2} + {x}^{4}}$
$\setminus \setminus \setminus \setminus \setminus \setminus = 2 \setminus \frac{1 - {x}^{2}}{{\left(1 - {x}^{2}\right)}^{2}}$
$\setminus \setminus \setminus \setminus \setminus \setminus = \frac{2}{1 - {x}^{2}}$

And differentiating again, and applying the chain rule, we gte

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 2 \left(- 1\right) {\left(1 - {x}^{2}\right)}^{- 2} \left(- 2 x\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{4 x}{1 - {x}^{2}} ^ 2 \setminus \setminus$ QED