Continuity and DifferentiabilityHard
Question
If y2 = p(x) is a polynomial of degree 3, then 2 
is equal to -
is equal to -Options
A.p″′(x) . p′(x)
B.p″(x) . p″′(x)
C.p(x) . p″(x)
D.none of these
Solution
y2 = p(x) ⇒ 2y
= p′(x)
⇒ 2y
+ 2 
= p″(x)
⇒ 2y
+ 2
= p″(x)
⇒ 4y3
+ (p′(x))2 = 2y2p″(x)
⇒ 4y3
= 2p(x) p″(x) - (p′(x))2
⇒ 2
=
[2p′(x)p″(x) + 2p(x)p″′(x) - 2p′(x)p″(x)]
⇒
= p(x)p″′(x)
= p′(x)⇒ 2y
+ 2 = p″(x)⇒ 2y
+ 2 = p″(x)⇒ 4y3
+ (p′(x))2 = 2y2p″(x)⇒ 4y3
= 2p(x) p″(x) - (p′(x))2⇒ 2
=
[2p′(x)p″(x) + 2p(x)p″′(x) - 2p′(x)p″(x)]⇒
= p(x)p″′(x)Create a free account to view solution
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