MHT CET · Maths · Differentiation
If \(\frac{x}{\sqrt{1+x}}+\frac{y}{\sqrt{1+y}}=0, x \neq y\), then \(\frac{d y}{d x}=0\)
- A \(\frac{1}{2}\)
- B 0
- C -1
- D 1
Answer & Solution
Correct Answer
(C) -1
Step-by-step Solution
Detailed explanation
(B)
\(\frac{x}{\sqrt{1+x}}+\frac{y}{\sqrt{1+y}}=0 \)
\( x \cdot(\sqrt{1+y})+y(\sqrt{1+x})=0 \quad \Rightarrow x \cdot \sqrt{1+y}\) \(=-y \sqrt{1+x}\)
Squaring both sides we get,
\(\begin{array}{l}
x^{2}(1+y)=y^{2}(1+x) \Rightarrow x^{2}+x^{2} y=y^{2}+x y^{2} \\
x^{2}-y^{2}=x y^{2}-x^{2} y \Rightarrow(x-y)(x+y)=-x y(x-y) \\
x+y=-x y \Rightarrow(1+x) y=-x \Rightarrow y=\frac{-x}{1+x}
\end{array}\)
Differentiating both sides w.r.t. \(x\), we get
\(\begin{aligned}
& \frac{d y}{d x}=-\frac{(1+x)(1)-(x)(1)}{(1+x)^{2}}=\frac{-1}{(1+x)^{2}} \\
\therefore &(1+x)^{2} \frac{d y}{d x}=-1
\end{aligned}\)
\(\frac{x}{\sqrt{1+x}}+\frac{y}{\sqrt{1+y}}=0 \)
\( x \cdot(\sqrt{1+y})+y(\sqrt{1+x})=0 \quad \Rightarrow x \cdot \sqrt{1+y}\) \(=-y \sqrt{1+x}\)
Squaring both sides we get,
\(\begin{array}{l}
x^{2}(1+y)=y^{2}(1+x) \Rightarrow x^{2}+x^{2} y=y^{2}+x y^{2} \\
x^{2}-y^{2}=x y^{2}-x^{2} y \Rightarrow(x-y)(x+y)=-x y(x-y) \\
x+y=-x y \Rightarrow(1+x) y=-x \Rightarrow y=\frac{-x}{1+x}
\end{array}\)
Differentiating both sides w.r.t. \(x\), we get
\(\begin{aligned}
& \frac{d y}{d x}=-\frac{(1+x)(1)-(x)(1)}{(1+x)^{2}}=\frac{-1}{(1+x)^{2}} \\
\therefore &(1+x)^{2} \frac{d y}{d x}=-1
\end{aligned}\)
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