# How do you find the second derivative of y=1/(t^2+1)?

Apr 8, 2017

$y ' ' = \frac{2 \left({t}^{2} - 1\right)}{{\left({t}^{2} + 1\right)}^{3}}$

#### Explanation:

$y = \frac{1}{{t}^{2} + 1}$

$y = {\left({t}^{2} + 1\right)}^{- 1}$

Chain rule:
$y ' = - {\left({t}^{2} + 1\right)}^{- 2} \left(2 t\right)$

$y ' = \frac{- 2 t}{{\left({t}^{2} + 1\right)}^{2}}$

Quotient rule and chain rule:
$y ' ' = \frac{{\left({t}^{2} + 1\right)}^{2} \left(- 2\right) - \left(- 2 t\right) \left(2\right) \left({t}^{2} + 1\right) \left(2 t\right)}{{\left({t}^{2} + 1\right)}^{4}}$

$y ' ' = \frac{\textcolor{red}{\left({t}^{2} + 1\right)} \left[- 2 \left({t}^{2} + 1\right) + 4 {t}^{2}\right]}{\textcolor{red}{{\left({t}^{2} + 1\right)}^{4}}}$

$y ' ' = \frac{- 2 {t}^{2} - 2 + 4 {t}^{2}}{{\left({t}^{2} + 1\right)}^{3}}$

$y ' ' = \frac{2 {t}^{2} - 2}{{\left({t}^{2} + 1\right)}^{3}}$

$y ' ' = \frac{2 \left({t}^{2} - 1\right)}{{\left({t}^{2} + 1\right)}^{3}}$

Apr 8, 2017

$\frac{{d}^{2} y}{{\mathrm{dt}}^{2}} = \frac{6 {t}^{2} - 2}{{t}^{2} + 1} ^ 3$

#### Explanation:

One way is to differentiate using the $\textcolor{b l u e}{\text{chain and product rules}}$

$\text{Express } \frac{1}{{t}^{2} + 1} = {\left({t}^{2} + 1\right)}^{-} 1$

differentiate using the $\textcolor{b l u e}{\text{chain rule}}$

$\text{given " y=f(g(x))" then}$

• dy/dx=f'(g(x))xxg'(x)

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dt}} = - {\left({t}^{2} + 1\right)}^{-} 2 \times \frac{d}{\mathrm{dt}} \left({t}^{2} + 1\right)$

$\textcolor{w h i t e}{\Rightarrow \frac{\mathrm{dy}}{\mathrm{dt}}} = - 2 t {\left({t}^{2} + 1\right)}^{-} 2$

$\text{differentiate "dy/dt" using the " color(blue)"product rule}$

$\text{given " f(x)=g(x).h(x)" then}$

• f'(x)=g(x)h'(x)+h(x)g'(x)

$g \left(t\right) = - 2 t \Rightarrow g ' \left(t\right) = - 2$

$h \left(t\right) = {\left({t}^{2} + 1\right)}^{-} 2 \Rightarrow h ' \left(t\right) = - 2 {\left({t}^{2} + 1\right)}^{-} 3 \left(2 t\right)$

$\textcolor{w h i t e}{X X X X X X X X X X X X X X} = - 4 t {\left({t}^{2} + 1\right)}^{-} 3$

$\Rightarrow \frac{{d}^{2} y}{{\mathrm{dt}}^{2}} = - 2 t . - 4 t {\left({t}^{2} + 1\right)}^{-} 3 + {\left({t}^{2} + 1\right)}^{-} 2 \left(- 2\right)$

$\textcolor{w h i t e}{\Rightarrow {d}^{2} \frac{y}{\mathrm{dt}} ^ 2} = 8 {t}^{2} {\left({t}^{2} + 1\right)}^{-} 3 - 2 {\left({t}^{2} + 1\right)}^{-} 2$

$\textcolor{w h i t e}{\Rightarrow {d}^{2} \frac{y}{\mathrm{dt}} ^ 2} = 2 {\left({t}^{2} + 1\right)}^{-} 3 \left(4 {t}^{2} - {t}^{2} - 1\right)$

$\textcolor{w h i t e}{\Rightarrow {d}^{2} \frac{y}{\mathrm{dt}} ^ 2} = \frac{6 {t}^{2} - 2}{{t}^{2} + 1} ^ 3$