Question
If x 1+y +y 1+x =0, then d y d x =
BITSAT
Mathematics
Differentiation
2024
BITSAT 2024 (Memory Based Paper 3)
BITSAT Mathematics Question (2024) — Solution
Options
- A. x+1 x
- B. 1 1+x
- C. -1 (1+x)^2
- D. x 1+x
Answer
C. -1 (1+x)^2
Step-by-step solution
aligned & Given, x 1+y +y 1+x =0 \\ & x 1+y =-y 1+x \\ & (x 1+y )^2=(-y 1+x )^2 aligned array lr & x^2(1+y)=y^2(1+x) \\ & x^2-y^2=y^2 x-x^2 y \\ & (x-y)(x+y)=x y(y-x) \\ & x+y=-x y \\ & x+y+x y=0 \\ & y= -x 1+x =-1+ 1 1+x array Differentiating w.r.t, x d y d x =- 1 (1+x)^2
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