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

Step 1: Understanding the Concept:
This is an implicit differentiation problem involving a recursive or self-referential equation. The variable $y$ appears on both sides, and it's inside a square root on one side.

Step 2: Key Formula or Approach:
To make differentiation easier, eliminate the square root by squaring both sides of the equation. Then, differentiate both sides implicitly with respect to $x$, applying the chain rule to terms involving $y$. Finally, group all terms containing $\frac{dy}{dx}$ and solve for it.

Step 3: Detailed Explanation:
The given equation is: \[ y = \sqrt{\tan x + y} \] 1. Square both sides to remove the radical: \[ y^2 = \tan x + y \] 2. Differentiate implicitly with respect to $x$: Apply the power rule and chain rule to $y^2$: $\frac{d}{dx}(y^2) = 2y \cdot \frac{dy}{dx}$. Differentiate the right side: $\frac{d}{dx}(\tan x) = \sec^2 x$, and $\frac{d}{dx}(y) = \frac{dy}{dx}$. Putting it together: \[ 2y \frac{dy}{dx} = \sec^2 x + \frac{dy}{dx} \] 3. Solve for $\frac{dy}{dx}$: Move all terms containing $\frac{dy}{dx}$ to the left side of the equation: \[ 2y \frac{dy}{dx} - \frac{dy}{dx} = \sec^2 x \] Factor out $\frac{dy}{dx}$ on the left side: \[ \frac{dy}{dx} (2y - 1) = \sec^2 x \] Divide both sides by $(2y - 1)$ to isolate $\frac{dy}{dx}$: \[ \frac{dy}{dx} = \frac{\sec^2 x}{2y - 1} \]

Step 4: Final Answer:
The derivative $\frac{dy}{dx}$ is $\frac{\sec^2 x}{2y - 1}$.