First slide

Derivatives

Question

$If y = \sqrt{\tan x + \sqrt{\tan x + \sqrt{\tan x + \dots \dots}}} then \frac{d y}{d x} =$

Moderate

A

$\frac{\cos^{2} x}{2 y - 1}$
B

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

$\frac{\tan x}{2 y - 1}$
D

$\frac{\cot x}{2 y - 1}$

Solution

$y = \sqrt{\tan x + y} \Rightarrow y^{2} = \tan x + y \Rightarrow 2 y \cdot \frac{d y}{d x} = \sec^{2} x + \frac{d y}{d x} \Rightarrow \frac{d y}{d x} = \frac{\sec^{2} x}{2 y - 1} .$

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