If y2(2−x)=x3, then dydx(1,1) is equal to:

Given: y2(2−x)=x3 ⇒(2−x)2ydydx+y2(−1)=3x2⇒(2−x)2ydydx=3x2+y2⇒dydx=3x2+y22y(2−x)⇒dydx(1,1)=2

If y2(2−x)=x3, then dydx(1,1) is equal to:

Given: y2(2−x)=x3 ⇒(2−x)2ydydx+y2(−1)=3x2⇒(2−x)2ydydx=3x2+y2⇒dydx=3x2+y22y(2−x)⇒dydx(1,1)=2

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If $y^{2} (2 - x) = x^{3}, then {(\frac{d y}{d x})}_{(1, 1)}$ is equal to:

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-2

$- \frac{3}{2}$

answer is A.

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Detailed Solution

$G i v e n : y^{2} (2 - x) = x^{3} \begin{array}{l} \Rightarrow (2 - x) 2 y \frac{d y}{d x} + y^{2} (- 1) = 3 x^{2} \\ \Rightarrow (2 - x) 2 y \frac{d y}{d x} = 3 x^{2} + y^{2} \\ \Rightarrow \frac{d y}{d x} = \frac{3 x^{2} + y^{2}}{2 y (2 - x)} \\ {\Rightarrow \frac{d y}{d x}|}_{(1, 1)} = 2 \end{array}$