TS EAMCET · Maths · Differentiation
If \(x \sqrt{1+y}+y \sqrt{1+x}=0\), then \(\frac{d y}{d x}\) is equal to
- A \(\frac{1}{(1+x)^2}\)
- B \(-\frac{1}{(1+x)^2}\)
- C \(\frac{1}{1+x^2}\)
- D \(\frac{1}{1-x^2}\)
Answer & Solution
Correct Answer
(B) \(-\frac{1}{(1+x)^2}\)
Step-by-step Solution
Detailed explanation
Given that \(x \sqrt{1+y}=-y \sqrt{1+x}\) On squaring both sides, we get \(\begin{array}{cc} & x^2(1+y)=y^2(1+x) \\ \Rightarrow & x^2-y^2+x^2 y-x y^2=0 \\ \Rightarrow & (x-y)(x+y)+x y(x-y)=0 \\ \Rightarrow & (x-y)(x+y+x y)=0 \end{array}\) \(x-y \neq 0\) because it does not…
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