Applications Of Derivatives Question 210
Question: If $ y^{2}=p(x) $ is a polynomial of degree three, then $ 2\frac{d}{dx}{ y^{3}.\frac{d^{2}y}{dx^{2}} } $ =
[IIT 1988; RPET 2000]
Options:
A) $ {p}’’’(x)+p’(x) $
B) $ {p}’’(x).{p}’’’(x) $
C) $ p(x).{p}’’’(x) $
D) Constant
Show Answer
Answer:
Correct Answer: C
Solution:
$ 2y\frac{dy}{dx}=p’(x)\Rightarrow 2\frac{dy}{dx}=\frac{{p}’(x)}{y} $
$ \Rightarrow 2\frac{d^{2}y}{dx^{2}}=\frac{y{p}’’(x)-{p}’(x){y}’}{y^{2}} $
therefore $ 2y^{3}\frac{d^{2}y}{dx^{2}}=y^{2}{p}’’(x)-y\frac{dy}{dx}{p}’(x) $
$ =p(x),{p}’’(x)-\frac{1}{2}{{{{p}’(x)}}^{2}} $
therefore $ 2\frac{d}{dx}( y^{3}\frac{d^{2}y}{dx^{2}} )={p}’(x){p}’’(x)+p(x){p}’’’(x)-{p}’(x){p}’’(x) $
$ =p(x){p}’’’(x) $ .
BETA
