Question

If $y=\frac{1}{x^2-a^2}$, then $\frac{d^{2}y}{d{x}^2}$ equals?

Answer 1

$y=\frac{1}{x^2-a^2}$

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

$\frac{d^{2}y}{d{x}^2}=\frac{(x^2-a^2{)}^2(-2)-(-2x\left).2\right(x^2-a^2\left)\right(2x)}{(x^{2-a^2}{)}^4}$

$\frac{d^{2}y}{d{x}^2}=\frac{-2(x^2-a^2{)}^2+8x^2(x^2-a^2)}{(x^{2-a^2}{)}^4}$

$\frac{d^{2}y}{d{x}^2}=\frac{-2(x^2-a^2)+8x^2}{(x^{2-a^2}{)}^3}$

$\frac{d^{2}y}{d{x}^2}=\frac{-2x^2+2a^2+8x^2}{(x^2-a^2{)}^3}=\frac{2a^2+6x^2}{(x^2-a^2{)}^3}$

$\frac{d^{2}y}{d{x}^2}=\frac{2(a^2+3x^2)}{(x^2-a^2{)}^3}$