X-1 - y - y-1 - x - 0Differentiate with respect to x -

Apply differentiation on both sides with respect to x √(()1+y)+y^!x2√(()1+y)+y2√(()1+x)+y^!√(()1+x) from solving this we get #160; y^!= (2(1+y)√(()1+x ...

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Byju's Answer

Standard XII

Mathematics

Chain Rule of Differentiation

x√1 + y + y√1...

Question

$x \sqrt{1 + y} + y \sqrt{1 + x} = 0$
Differentiate with respect to $x$ :-

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Solution

apply differentiation on both sides with respect to x
$\sqrt{(} 1 + y) + \frac{y^{!} x}{2 \sqrt{(} 1 + y)} + \frac{y}{2 \sqrt{(} 1 + x)} + y^{!} \sqrt{(} 1 + x)$
from solving this we get
$y^{!} = \frac{- (2 (1 + y) \sqrt{(} 1 + x) + y \sqrt{(} 1 + y))}{\sqrt{(} 1 + x) + 2 (1 + x) \sqrt{(} 1 + y)}$

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x√1+y+y√1+x=0Differentiate with respect to x :-

apply differentiation on both sides with respect to x√(1+y)+y!x2√(1+y)+y2√(1+x)+y!√(1+x)from solving this we get y!=−(2(1+y)√(1+x)+y√(1+y))√(1+x)+2(1+x)√(1+y)

$x \sqrt{1 + y} + y \sqrt{1 + x} = 0$
Differentiate with respect to $x$ :-

apply differentiation on both sides with respect to x
$\sqrt{(} 1 + y) + \frac{y^{!} x}{2 \sqrt{(} 1 + y)} + \frac{y}{2 \sqrt{(} 1 + x)} + y^{!} \sqrt{(} 1 + x)$
from solving this we get
$y^{!} = \frac{- (2 (1 + y) \sqrt{(} 1 + x) + y \sqrt{(} 1 + y))}{\sqrt{(} 1 + x) + 2 (1 + x) \sqrt{(} 1 + y)}$