Question

If $ y = \sqrt{\sin x + y} $, then find $ \frac{dy}{dx} $ at $ x = 0, y = 1 $.

• A. 0
• B. 1
• C. 2
• D. -1

Hint:

When performing implicit differentiation, make sure to differentiate both sides carefully and remember to apply the chain rule when necessary.

Answer 1

The Correct Option is B

We are given the equation: \[ y = \sqrt{\sin x + y} \] To find \( \frac{dy}{dx} \), we will differentiate both sides with respect to \( x \) using implicit differentiation.
Step 1: Differentiate both sides
Start by differentiating the left-hand side and the right-hand side of the equation: \[ \frac{d}{dx}\left(y\right) = \frac{d}{dx}\left(\sqrt{\sin x + y}\right) \] The left-hand side becomes: \[ \frac{dy}{dx} \] Now differentiate the right-hand side using the chain rule: \[ \frac{d}{dx}\left(\sqrt{\sin x + y}\right) = \frac{1}{2\sqrt{\sin x + y}}\left(\cos x + \frac{dy}{dx}\right) \] Thus, the equation becomes: \[ \frac{dy}{dx} = \frac{1}{2\sqrt{\sin x + y}}\left(\cos x + \frac{dy}{dx}\right) \]
Step 2: Solve for \( \frac{dy}{dx} \)
Now, substitute \( x = 0 \) and \( y = 1 \) into the equation. At \( x = 0 \): \[ y = \sqrt{\sin(0) + y} = \sqrt{0 + 1} = 1 \] Substitute \( y = 1 \) and \( x = 0 \) into the equation for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{1}{2\sqrt{0 + 1}}\left(\cos(0) + \frac{dy}{dx}\right) \] \[ \frac{dy}{dx} = \frac{1}{2}(1 + \frac{dy}{dx}) \] Now simplify: \[ \frac{dy}{dx} = \frac{1}{2} + \frac{1}{2} \frac{dy}{dx} \] \[ \frac{dy}{dx} - \frac{1}{2} \frac{dy}{dx} = \frac{1}{2} \] \[ \frac{1}{2} \frac{dy}{dx} = \frac{1}{2} \] Thus: \[ \frac{dy}{dx} = 1 \] Therefore, the value of \( \frac{dy}{dx} \) at \( x = 0 \) and \( y = 1 \) is \( 1 \).