Question

If $y=\frac{1}{2x^2+3x+1}$ then $\frac{d^{2}y}{d{x}^2}$ at $x=-2$ is :

• A. $\frac{110}{27}$
• B. $-\frac{38}{27}$
• C. $\frac{27}{38}$
• D. $none$

Answer 1

Answer: (A)

$\frac{110}{27}$

$y\frac{1}{2x^2+3x+1}$

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

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

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

$=\frac{d^{2}y}{d{x}^2}=\left[-4{\left(2x^2+3x+1\right)}^{-2}\left(4x+3\right)\times \left(-2\right){\left(2x^2+3x+1\right)}^{-3}{\left(4x+3\right)}^2\right]$

$=\frac{d^{2}y}{d{x}^2}=-\left[4\left(4x+3\right){\left(2x^2+3x+1\right)}^{-2}\times -2{\left(4x+3\right)}^2{\left(2x^2+3x+1\right)}^{-3}\right]$

$={\left(\frac{d^{2}y}{d{x}^2}\right)}_{x=-2}=-\left[4\left(-5\right)\left(\frac{1}{9}\right)-2\left(25\right)\left(\frac{1}{27}\right)\right]$

$=-\left[\frac{-20}{9}-\frac{50}{27}\right]$

$=-\left[\frac{-60-50}{27}\right]$

$=\frac{110}{27}$