Q60 / 125 · BITSAT 2020 · Mathematics

If $x y+y^{2}=\tan x+y,$ then find $\frac{d y}{d x}$ is

  • A
    $\frac{\sec ^{2} x}{x+2 y}$
  • Correct answer
    B
    $\frac{\sec ^{2} x-y}{(x+2 y-1)}$
  • C
    $(x+2 y-1) \sec ^{2} x$
  • D
    $\sec ^{2} x \cdot y$

Solution

The given relation is $x y+y^{2}=\tan x+y$. Differentiating both sides with respect to $x$, we get $\frac{d}{d x}(x y)+\frac{d}{d x}\left(y^{2}\right)=\frac{d}{d x}(\tan x)+\frac{d y}{d x}$ or $\left[y \cdot 1+x \cdot \frac{d y}{d x}\right]+2 y \frac{d y}{d x}=\sec ^{2} x+\frac{d y}{d x}$ or $(x+2 y-1) \frac{d y}{d x}=\sec ^{2} x-y$ $\therefore \frac{d y}{d x}=\frac{\sec ^{2} x-y}{(x+2 y-1)}$

Hence, the correct answer is option: 'B'.

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