Question
If x y+y^ 2 = x+y, then find d y d x is
BITSAT
Mathematics
Differentiation
2020
BITSAT 2020
BITSAT Mathematics Question (2020) — Solution
Options
- A. ^ 2 x x+2 y
- B. ^ 2 x-y (x+2 y-1)
- C. (x+2 y-1) ^ 2 x
- D. ^ 2 x y
Answer
B. ^ 2 x-y (x+2 y-1)
Step-by-step solution
The given relation is x y+y^ 2 = x+y. Differentiating both sides with respect to x, we get d d x (x y)+ d d x (y^ 2 )= d d x ( x)+ d y d x or [y 1+x d y d x ]+2 y d y d x = ^ 2 x+ d y d x or (x+2 y-1) d y d x = ^ 2 x-y d y d x = ^ 2 x-y (x+2 y-1)
Related: Mathematics — Differentiation · All PYQ Banks