Question

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

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

Answer 1

The correct option is A $\frac{38}{27}$

Given, $2x^2+3x+1=\frac{1}{y}$
$(4x+3)=\frac{-1}{y^2}.y^{\prime }$
$y^2(4x+3)=-y^{\prime }$
$y^2.4+2y{y}^{\prime }(4x+3)=-y^{\prime \prime }$
$y{|}_{x=-2}=\frac{1}{3},y^2{|}_{x=-2}=\frac{1}{9}and{y}^{\prime }{|}_{x=-2}=\frac{5}{9}$
$\frac{4}{9}-\frac{50}{27}=-y^{\prime \prime }$
$y^{\prime \prime }=\frac{38}{27}$