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If $x y+y^{2}=\tan x+y,$ then find $\frac{d y}{d x}$ is
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Verified Answer
The correct answer is:
$\frac{\sec ^{2} x-y}{(x+2 y-1)}$
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 $(x+2 y-1) \frac{d y}{d x}=\sec ^{2} x-y$
$\therefore \frac{d y}{d x}=\frac{\sec ^{2} x-y}{(x+2 y-1)}$
$\frac{d}{d x}(x y)+\frac{d}{d x}\left(y^{2}\right)=\frac{d}{d x}(\tan x)+\frac{d y}{d x}$
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 $(x+2 y-1) \frac{d y}{d x}=\sec ^{2} x-y$
$\therefore \frac{d y}{d x}=\frac{\sec ^{2} x-y}{(x+2 y-1)}$
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