Question

If $y^2=a{x}^2+bx+c$, where a, b, c are constants, then $y^3\frac{d^{2}y}{d{x}^2}$ is equal to.

• A. $\frac{-1}{4}$
• B. $\frac{-1}{6}$
• C. $\frac{-3}{4}$
• D. $\frac{-1}{8}$

Answer 1

The correct option is A $\frac{-1}{4}$

$y^2=a{x}^2+bx+c$

$y=(a{x}^2+6x+c{)}^{\frac{1}{2}}$

$\frac{dy}{dx}=\frac{1}{2}(a{x}^2+6x+c{)}^{\frac{-1}{2}}$

$\frac{d^{2}y}{dx}=\frac{-1}{4}(a{x}^2+6x+c{)}^{\frac{-3}{2}}$

$y^3=(a{x}^2+6x+c{)}^{\frac{3}{2}}$

$y^3\frac{d^{2}y}{d{x}^2}=(a{x}^2+6x+c{)}^{\frac{3}{2}}\times \frac{-1}{4}(a{x}^2+6x+c{)}^{\frac{3}{2}}$

$y^3\frac{d^{2}y}{d{x}^2}=\frac{-1}{4}$