Question

If $xy+y^2=\tan x+y$, then $\frac{dy}{dx}$ is equal to

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

Answer 1

Answer: (A)

$\frac{{\sec }^{2}x-y}{(x+2y-1)}$

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