Let f ( x ) = sin ( 2 sin − 1 x ) for x ∈ ( − 1 , 1 ) . Consider the values of its derivatives at x = 0 . List - I List - II (P) f ′ ( 0 ) (1) 0 (Q) f ′′ ( 0 ) (2) 2 (R) f ′′′ ( 0 ) (3) -6 (S) f ( 4 )

Question

Let f ( x ) = sin ( 2 sin − 1 x ) for x ∈ ( − 1 , 1 ) . Consider the values of its derivatives at x = 0 . List - I List - II (P) f ′ ( 0 ) (1) 0 (Q) f ′′ ( 0 ) (2) 2 (R) f ′′′ ( 0 ) (3) -6 (S) f ( 4 ) ( 0 ) (4) -30 (5) 1 (6) 4

TestHub · Accepted Answer

Let y = f ( x ) = sin ( 2 sin − 1 x ) . Differentiating, y ′ = cos ( 2 sin − 1 x ) ⋅ 1 − x 2 ​ 2 ​ . This leads to the differential equation ( 1 − x 2 ) y ′′ − x y ′ + 4 y = 0 . Applying Leibniz theorem for the n -th derivative and evaluating at x = 0 , we get y ( n + 2 ) ( 0 ) = ( n 2 − 4 ) y ( n ) ( 0 ) . We have y ( 0 ) = 0 and y ′ ( 0 ) = 2 . Using the recurrence: y ′′ ( 0 ) = ( 0 2 − 4 ) y ( 0 ) = 0 . y ′′′ ( 0 ) = ( 1 2 − 4 ) y ′ ( 0 ) = − 3 ( 2 ) = − 6 . y ( 4 ) ( 0 ) = ( 2 2 − 4 ) y ′′ ( 0 ) = 0 ( 0 ) = 0 . Thus, P-(2), Q-(1), R-(3), S-(1).

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Doubts & Discussion

List - I	List - II
(P) $f^{'} (0)$	(1) 0
(Q) $f^{''} (0)$	(2) 2
(R) $f^{'''} (0)$	(3) -6
(S) $f^{(4)} (0)$	(4) -30
	(5) 1
	(6) 4