Question

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

• A. $\frac{xy}{x-x^2-y^2}$
• B. $\frac{y}{x-x^2-y^2}$
• C. $\frac{\tan y}{1-x{\sec }^{2}y}$
• D. $\frac{y}{1-x{\sec }^{2}y}$

Answer 1

Answer: (B), (C)

$\frac{\tan y}{1-x{\sec }^{2}y}$

We have,
$y=x\tan y\cdots \left(1\right)$
Differentiating both sides w.r.t. $x,$ we get
$\frac{dy}{dx}=\tan y+x{\sec }^{2}y\cdot \frac{dy}{dx}$
$\frac{dy}{dx}[1-x{\sec }^{2}y]=\tan y$
$\frac{dy}{dx}=\frac{\tan y}{1-x{\sec }^{2}y}$
$=\frac{\frac{y}{x}}{1-x(1+\frac{y^2}{x^2})}\text{(Using (1))}$
$=\frac{y}{x-x^2-y^2}$