# What is the second derivative of f(x)=(x-1)/(x^2+1)?

Mar 6, 2017

$f ' ' \left(x\right) = \frac{2 \left(1 + x\right) \left(1 - 4 x + {x}^{2}\right)}{1 + {x}^{2}} ^ 3.$

#### Explanation:

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

Using the Quotient Rule for Diffn., we get, the first deri.,

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

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

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

Rediff.ing $f ' \left(x\right)$ to get second deri.,

$f ' ' \left(x\right) = \frac{{\left(1 + {x}^{2}\right)}^{2} \frac{d}{\mathrm{dx}} \left(1 + 2 x - {x}^{2}\right) - \left(1 + 2 x - {x}^{2}\right) \frac{d}{\mathrm{dx}} \left(1 + 2 {x}^{2} + {x}^{4}\right)}{{\left({x}^{2} + 1\right)}^{2}} ^ 2$

$= \frac{{\left(1 + {x}^{2}\right)}^{2} \left(2 - 2 x\right) - \left(1 + 2 x - {x}^{2}\right) \left(4 x + 4 {x}^{3}\right)}{{x}^{2} + 1} ^ 4$

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

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

$= \frac{2 \left(1 + {x}^{2}\right) \left\{1 + {x}^{2} - x - {x}^{3} - 2 x - 4 {x}^{2} + 2 {x}^{3}\right\}}{{x}^{2} + 1} ^ 4$

$\therefore f ' ' \left(x\right) = \frac{2 \left(1 - 3 x - 3 {x}^{2} + {x}^{3}\right)}{1 + {x}^{2}} ^ 3.$

$= \frac{2 \left\{\left(1 + {x}^{3}\right) - 3 x \left(1 + x\right)\right\}}{1 + {x}^{2}} ^ 3$

$= \frac{2 \left\{\left(1 + x\right) \left(1 - x + {x}^{2}\right) - 3 x \left(1 + x\right)\right\}}{1 + {x}^{2}} ^ 3$

$= \frac{2 \left(1 + x\right) \left(1 - x + {x}^{2} - 3 x\right)}{1 + {x}^{2}} ^ 3$

$\Rightarrow f ' ' \left(x\right) = \frac{2 \left(1 + x\right) \left(1 - 4 x + {x}^{2}\right)}{1 + {x}^{2}} ^ 3$

Enjoy Maths.!