# How do you find (d^2y)/(dx^2) given y=(2x+5)^(-1/2)?

Nov 12, 2016

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{3}{2 x + 5} ^ \left(\frac{5}{2}\right)$

#### Explanation:

To find the $\textcolor{b l u e}{\text{first derivative}} \frac{\mathrm{dy}}{\mathrm{dx}}$ differentiate using the $\textcolor{b l u e}{\text{chain rule}}$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dx}}} \textcolor{w h i t e}{\frac{2}{2}} |}}} \to \left(A\right)$

let $u = 2 x + 5 \Rightarrow \frac{\mathrm{du}}{\mathrm{dx}} = 2$

and $y = {u}^{- \frac{1}{2}} \Rightarrow \frac{\mathrm{dy}}{\mathrm{du}} = - \frac{1}{2} {u}^{- \frac{3}{2}}$

substitute into (A) changing u back to x.

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{2} {u}^{- \frac{3}{2}} \times 2 = - {\left(2 x + 5\right)}^{- \frac{3}{2}}$

To find the $\textcolor{b l u e}{\text{second derivative }} \frac{{d}^{2} y}{{\mathrm{dx}}^{2}}$ repeat the process above on $\frac{\mathrm{dy}}{\mathrm{dx}}$ using the $\textcolor{b l u e}{\text{chain rule}}$

$u = 2 x + 5 \Rightarrow \frac{\mathrm{du}}{\mathrm{dx}} = 2$

$y = - {u}^{- \frac{3}{2}} \Rightarrow \frac{\mathrm{dy}}{\mathrm{du}} = \frac{3}{2} {u}^{- \frac{5}{2}}$

$\Rightarrow \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 3 {\left(2 x + 5\right)}^{- \frac{5}{2}} = \frac{3}{2 x + 5} ^ \left(\frac{5}{2}\right)$