Mathematics - Differentiation Question with Solution | TestHub

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