Limits, Continuity and Differentiability
80299
If \(\frac{x}{\sqrt{1+x}}+\frac{y}{\sqrt{1+y}}=0, x \neq y\), then \((1+x)^{2} \frac{d y}{d x}=\)
1 0
2 -1
3 \(\frac{1}{2}\)
4 1
Explanation:
(B) : Given,
\(\frac{x}{\sqrt{1+x}}+\frac{y}{\sqrt{1+y}}=0, x \neq y\)
\(\Rightarrow \quad \frac{x}{\sqrt{1+x}}=-\frac{y}{\sqrt{1+y}}=0\)
On squaring both sides,
\(\Rightarrow \frac{x^{2}}{(1+x)}=\frac{y^{2}}{(1+y)}\)
\(\Rightarrow x^{2}(1+y)=y^{2}(1+x)\)
\(\Rightarrow x^{2}+x^{2} y=y^{2}+y^{2} x\)
\(\Rightarrow x^{2} y-y^{2} x=y^{2}-x^{2}\)
\(\Rightarrow x y(x-y)=-(x-y)(x+y)\)
\(\Rightarrow x y=-(x+y)\)
\(x y+y=-x\)
\(\Rightarrow \quad y=-\frac{x}{1+x}\)
Differentiating w.r.to \(\mathrm{x}\).
\(\frac{d y}{d x}=-\frac{(1+x) \cdot 1-x \cdot 1}{(1+x)^{2}} \Rightarrow \frac{d y}{d x}=-\frac{-1}{(1+x)^{2}}\)
\((1+x)^{2} \frac{d y}{d x}=-1\)
