Limit Continuity And Differentiability Ques 13
- If $y^2=P(x)$ is a polynomial of degree $3$ , then $2 \frac{d}{d x}\left(y^3 \frac{d^2 y}{d x^2}\right)$ equals
$(1988,2 M)$
(a) $P^{\prime \prime \prime}(x)+P^{\prime}(x)$
(b) $P^{\prime \prime}(x) \cdot P^{\prime \prime}(x)$
(c) $P(x)\cdot P^{ \prime\prime\prime}(x)$
(d) a constant
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Answer:
Correct Answer: 13.(c)
Solution: (c) Since, $y^2=P(x)$
On differentiating both sides, we get
$ 2 y y_1=P^{\prime}(x), $
Again, differentiating, we get
$2 y y_2+2 y_1^2 =P^{\prime \prime}(x) $
$\Rightarrow \quad 2 y^3 y_2+2 y^2 y_1^2 =y^2 P^{\prime \prime}(x) $
$\Rightarrow \quad 2 y^3 y_2 =y^2 P^{\prime \prime}(x)-2\left(y y_1\right)^2 $
$ \Rightarrow \quad 2 y^3 y_2 =P(x) \cdot P^{\prime \prime}(x)-\frac{\left\{P^{\prime}(x)\right\}^2}{2}$
Again, differentiating, we get
$ \begin{aligned} & 2 \frac{d}{d x}\left(y^3 y_2\right)=P^{\prime}(x) \cdot P^{\prime \prime}(x)+P(x) \cdot P^{\prime \prime \prime}(x) -\frac{2 P^{\prime}(x) \cdot P^{\prime \prime}(x)}{2} \\ & \Rightarrow \quad 2 \frac{d}{d x}\left(y^3 y_2\right)=P(x) \cdot P^{\prime \prime}(x) \\ & \Rightarrow \quad 2 \frac{d}{d x}\left(y^3 \cdot \frac{d^2 y}{d x^2}\right)=P(x) \cdot P^{\prime \prime \prime}(x) \\ \end{aligned} $
BETA
