If \( y^{3}-y-2x=0 \), then \( \displaystyle\frac{d^{2}y}{dx^{2}} \) is equal to:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{-24y}{(3y^{2}-1)^{3}} \)
Differentiating \( y^{3}-y-2x=0 \) with respect to \( x \):
\( \displaystyle (3y^{2}-1)\frac{dy}{dx}=2 \)
\( \displaystyle \frac{dy}{dx}=\frac{2}{3y^{2}-1} \)
Differentiating again, with the chain rule on \( 3y^{2}-1 \):
\( \displaystyle \frac{d^{2}y}{dx^{2}}=-\frac{2\cdot 6y}{(3y^{2}-1)^{2}}\cdot\frac{dy}{dx} \)
Substituting \( \frac{dy}{dx} \):
\( \displaystyle \frac{d^{2}y}{dx^{2}}=-\frac{12y}{(3y^{2}-1)^{2}}\cdot\frac{2}{3y^{2}-1} \)
\( \displaystyle =-\frac{24y}{(3y^{2}-1)^{3}} \)