Continuity and Differentiability Question - CBSE

Question

If $x \sqrt{1 + y} + y \sqrt{1 + x} = 0$, for $-1 < x < 1$, prove that $\dfrac{dy}{dx} = -\dfrac{1}{(1 + x)^2}$.

Accepted Answer

Squaring: $x^2(1+y) = y^2(1+x) \Rightarrow x^2 - y^2 = xy(y - x)$. [1.5 Marks]
Factorize and divide by $(x-y) \Rightarrow y(1+x) = -x \Rightarrow y = -\dfrac{x}{1+x}$. [1.5 Marks]
$\dfrac{dy}{dx} = -\dfrac{(1+x) - x}{(1+x)^2} = -\dfrac{1}{(1+x)^2}$. Proved! [2.0 Marks]

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🎯 Official CBSE Marking Scheme:
Squaring and simplifying implicit radical equation: 1.5 Marks
Expressing $y$ explicitly as a function of $x$: 1.5 Marks
Differentiating to obtain $-1/(1+x)^2$: 2.0 Marks

Question:

Step-by-Step Solution

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