Question

lf $f\left(x\right)=\frac{x^3}{x^2-1}$ then $f^{{}^{\prime \prime \prime }}\left(0\right)=$

• A. $-(3!)$
• B. $0$
• C. $1$
• D. $3!$

Answer 1

The correct option is A $-(3!)$

$(x^2-1)f\left(x\right)=x^3$
$2xf\left(x\right)+(x^2-1)f^{\prime }\left(x\right)=3x^2$
$2f\left(x\right)+2x{f}^{\prime }\left(x\right)+2x{f}^{\prime }\left(x\right)+(x^2-1)f^{\prime \prime }\left(x\right)=6x$
$2f^{\prime }\left(x\right)+4f^{\prime }\left(x\right)+6x{f}^{\prime \prime }\left(x\right)+(x^2-1)f^{\prime \prime \prime }\left(x\right)=6$
$f^{\prime \prime \prime }\left(x\right)=-6$