Question

If $y = \sqrt{\tan x + y}$, then $\frac{dy}{dx} = $

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

Hint:

For infinite nested root problems of the form $y = \sqrt{f(x) + \sqrt{f(x) + \dots}}$, which is equivalent to $y = \sqrt{f(x) + y}$, the derivative is always given by the standard shortcut formula: $y' = \frac{f'(x)}{2y - 1}$. Here $f(x) = \tan x$, so $y' = \frac{\sec^2 x}{2y-1}$.

Answer 1

The Correct Option is B

To solve the problem, we need to find the derivative of \( y = \sqrt{\tan x + y} \) with respect to \( x \).

First, let's square both sides to eliminate the square root:

\(y^2 = \tan x + y\)

Rearrange this equation:

\(y^2 - y = \tan x\)

To find \( \frac{dy}{dx} \), we differentiate both sides of the equation with respect to \( x \). Use implicit differentiation:

• The derivative of \( y^2 \) with respect to \( x \) is \(2y \frac{dy}{dx}\).
• The derivative of \( y \) with respect to \( x \) is \(\frac{dy}{dx}\).
• The derivative of \( \tan x \) with respect to \( x \) is \(\sec^2 x\).

Applying these, we get:

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

Factor out \(\frac{dy}{dx}\) from the left side:

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

Solve for \(\frac{dy}{dx}\) by dividing both sides by \((2y - 1)\):

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

Therefore, the correct answer is:

\(\frac{sec^2 x}{2y - 1}\)

This matches the given option B. Hence, the derivative of the function is:

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