Second derivative via implicit differentiation

A question that requires computing or evaluating d²y/dx² for an implicitly defined curve, possibly at a specific point.

jee-advanced 2008 Q18 View
Consider the functions defined implicitly by the equation $y ^ { 3 } - 3 y + x = 0$ on various intervals in the real line. If $x \in ( - \infty , - 2 ) \cup ( 2 , \infty )$, the equation implicitly defines a unique real valued differentiable function $y = f ( x )$. If $x \in ( - 2,2 )$, the equation implicitly defines a unique real valued differentiable function $y = g ( x )$ satisfying $g ( 0 ) = 0$.
If $f ( - 10 \sqrt { 2 } ) = 2 \sqrt { 2 }$, then $f ^ { \prime \prime } ( - 10 \sqrt { 2 } ) =$
(A) $\frac { 4 \sqrt { 2 } } { 7 ^ { 3 } 3 ^ { 2 } }$
(B) $- \frac { 4 \sqrt { 2 } } { 7 ^ { 3 } 3 ^ { 2 } }$
(C) $\frac { 4 \sqrt { 2 } } { 7 ^ { 3 } 3 }$
(D) $- \frac { 4 \sqrt { 2 } } { 7 ^ { 3 } 3 }$
jee-main 2018 Q81 View
If $x ^ { 2 } + y ^ { 2 } + \sin y = 4$, then the value of $\frac { d ^ { 2 } y } { d x ^ { 2 } }$ at the point $( - 2,0 )$ is :
(1) $-34$
(2) 4
(3) $-2$
(4) $-32$
jee-main 2018 Q81 View
If $x ^ { 2 } + y ^ { 2 } + \sin y = 4$, then the value of $\frac { d ^ { 2 } y } { d x ^ { 2 } }$ at the point $( - 2,0 )$ is
(1) - 34
(2) - 32
(3) - 2
(4) 4
jee-main 2020 Q63 View
If $y ^ { 2 } + \log _ { e } \left( \cos ^ { 2 } x \right) = y , \quad x \in \left( - \frac { \pi } { 2 } , \frac { \pi } { 2 } \right)$ then:
(1) $y \prime \prime ( 0 ) = 0$
(2) $| y \prime ( 0 ) | + | y \prime \prime ( 0 ) | = 1$
(3) $| y \prime \prime ( 0 ) | = 2$
(4) $| y \prime ( 0 ) | + | y \prime \prime ( 0 ) | = 3$
jee-main 2022 Q86 View
For the curve $C : \left( x ^ { 2 } + y ^ { 2 } - 3 \right) + \left( x ^ { 2 } - y ^ { 2 } - 1 \right) ^ { 5 } = 0$, the value of $3 y ^ { \prime } - y ^ { 3 } y ^ { \prime \prime }$, at the point $( \alpha , \alpha ) , \alpha > 0$, on $C$, is equal to $\_\_\_\_$ .