Question

If $y=\sqrt{\tan x+\sqrt{\tan x+\sqrt{\tan x+\cdots \text{to}\infty }}}$, then $\frac{dy}{dx}=\frac{p{\sec }^{2}x}{qy+r}$, $p,q,r\in N$. Find the minimum positive value of $p+q+r$.

Answer 1

Square both sides the given equation,

$\Rightarrow y^2=\tan x+y$

Now differentiate both sides with respect to $x$

$\Rightarrow 2y\frac{dy}{dx}={\sec }^{2}x+\frac{dy}{dx}$

$\Rightarrow (2y-1)\frac{dy}{dx}={\sec }^{2}x$

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

So, $p=1,q=2,r=-1$

$\therefore p+q+r=2$