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BITSAT 2024
Exam
BITSAT
Year
2024
Subject
Mathematics
Chapter
Differentiation

BITSAT 2024 · Mathematics · Differentiation

If x 1+y +y 1+x =0 , then (d y)/(d x)=

If x1+y+y1+x=0x \sqrt{1+y}+y \sqrt{1+x}=0, then dydx=\frac{d y}{d x}=

Options

  1. Ax+1x\frac{x+1}{x}
  2. B11+x\frac{1}{1+x}
  3. C−1(1+x)2\frac{-1}{(1+x)^2}
  4. Dx1+x\frac{x}{1+x}

Correct answer

C. −1(1+x)2\frac{-1}{(1+x)^2}

Step-by-step solution

 Given, x1+y+y1+x=0⇒x1+y=−y1+x⇒(x1+y)2=(−y1+x)2\begin{aligned} & \text{ Given, } x \sqrt{1+y}+y \sqrt{1+x}=0 \\ & \Rightarrow \quad x \sqrt{1+y}=-y \sqrt{1+x} \\ & \Rightarrow \quad(x \sqrt{1+y})^2=(-y \sqrt{1+x})^2 \end{aligned}⇒x2(1+y)=y2(1+x)⇒x2−y2=y2x−x2y⇒(x−y)(x+y)=xy(y−x)⇒x+y=−xy⇒x+y+xy=0⇒y=−x1+x=−1+11+x\begin{array}{lr} \Rightarrow & x^2(1+y)=y^2(1+x) \\ \Rightarrow & x^2-y^2=y^2 x-x^2 y \\ \Rightarrow & (x-y)(x+y)=x y(y-x) \\ \Rightarrow & x+y=-x y \\ \Rightarrow & x+y+x y=0 \\ \Rightarrow & y=\frac{-x}{1+x}=-1+\frac{1}{1+x} \end{array} Differentiating w.r.t, xx\frac{d y}{d x}=−-\frac{1}{(1+x)^2}$
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