Q

# If x 1+ y raised to 1 by 2 + y 1+ x raised to 1 by 2 equals 0 , for , – 1 < x < 1, prove that

Q14  If $x \sqrt {1+ y }+ y \sqrt { 1+ x } = 0 \: \: for \: \: , -1 < x < 1 \: \:prove \: \: that \: \frac{dy}{dx} = -\frac{1}{(1+x)^2}$

Views

Given function is
$x \sqrt {1+ y }+ y \sqrt { 1+ x } = 0$
$x\sqrt{1+y} = - y\sqrt{1+x}$
Now, squaring both sides
$(x\sqrt{1+y})^2 = (- y\sqrt{1+x})^2\\ x^2(1+y)=y^2(1+x)\\ x^2+x^2y=y^2x+y^2\\ x^2-y^2=y^2x-x^2y\\ (x-y)(x+y) = -xy(x-y) \\ x+y =-xy\\ y = \frac{-x}{1+x}$
Now, differentiate w.r.t. x is
$\frac{dy}{dx} = \frac{d(\frac{-x}{1+x})}{dx}= \frac{-1.(1+x)-(-x).(1)}{(1+x)^2}= \frac{-1}{(1+x)^2}$
Hence proved

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