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Implicit Differentiation

If x√1+y+y√...

Question

If $x \sqrt{1 + y} + y \sqrt{1 + x} = 0$ , for $- 1 < x < 1$ , prove that $\frac{d y}{d x} = \frac{1}{{(1 + x)}^{2}}$

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Solution

Given, $x \sqrt{1 + y} + y \sqrt{1 + x} = 0$
If we differentiate this function with respect to $x$ , then we get

$\sqrt{1 + y} + \frac{x}{2 \sqrt{1 + y}} \frac{d y}{d x} + \frac{y}{2 \sqrt{1 + x}} + \frac{d y}{d x} \sqrt{1 + x} = 0$
Taking out $\frac{d y}{d x}$ one side, we get $\frac{d y}{d x} = \frac{1}{{(1 + x)}^{2}}$

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Similar questions

x \sqrt{1 + y} + y \sqrt{1 + x} = 0

\frac{d y}{d x} = - \frac{1}{(1 + x)^{2}}

x \sqrt{1 + y} + y \sqrt{1 + x} = 0, - 1 < x < 1, x \neq y

then prove that

\frac{d y}{d x} = - \frac{1}{{(1 + x)}^{2}}

x \sqrt{1 + y} + y \sqrt{1 + x} = 0

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\frac{d y}{d x} = - \frac{1}{(x + 1)^{2}}

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x \sqrt{1 - y^{2}} + y \sqrt{1 - x^{2}} = 1

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