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Question

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


Solution

Square both sides the given equation, 
$$\Rightarrow y^2 = \tan x+y$$
Now differentiate both sides with respect to $$x$$
$$\Rightarrow 2y\dfrac{dy}{dx}=\sec^2x+\dfrac{dy}{dx}$$
$$\Rightarrow  (2y-1)\dfrac{dy}{dx}=\sec^2x$$
$$\therefore \dfrac{dy}{dx}=\dfrac{\sec^2x}{2y-1}$$
So, $$p=1, q = 2, r = -1$$
$$\therefore p+q+r=2$$

Mathematics

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