Question

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

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

Answer 1

The correct option is B $\frac{y}{x-x^2-y^2}$

$y=x\tan y$ $⟶\left(1\right)$

diff. both sides w.r.t. $x$

$\frac{dy}{dx}=1\cdot \tan y+x{\sec }^{2}y\cdot \frac{dy}{dx}$

$\frac{dy}{dx}[1-x{\sec }^{2}y]=\tan y$

$\frac{dy}{dx}=\frac{\tan y}{1-x{\sec }^{2}y}$ $⟶\left(2\right)$

from eqn (1) $\tan y=\frac{y}{x}$

${\tan }^{2}y=\frac{y^2}{x^2}$

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

${\sec }^{2}y=1+\frac{y^2}{x^2}$

But value of ${\sec }^{2}y$ in eqn (2)

$\frac{dy}{dx}=\frac{\tan y}{1-x[1+\frac{y^2}{x^2}]}=\frac{\tan y}{1-x\left[\frac{x^2+y^2}{x^2}\right]}$

$=\frac{\tan y}{1-\left(\frac{x^2+y^2}{x}\right)}=\frac{x\tan y}{x-x^2-y^2}$

$\frac{dy}{dx}=\frac{y}{x-x^2-y^2}$ $\left[fromeqn\right(1\left)\right]$