Question

If \[ x^4 + 2\sqrt{y} + 1 = 3, \] then \( \frac{dy}{dx} \) at \( (1,0) \) is equal to

• A. \( 4 \)
• B. \( 2 \)
• C. \( -4 \)
• D. \( -2 \)
• E. \( -\frac{1}{8} \)

Hint:

For implicit differentiation, differentiate both sides carefully and use chain rule when required.

Answer 1

The Correct Option is C

Differentiating both sides with respect to \( x \): \[ \frac{d}{dx} \left( x^4 + 2\sqrt{y} + 1 \right) = \frac{d}{dx} (3) \] \[ 4x^3 + 2 \cdot \frac{1}{2} y^{-1/2} \cdot \frac{dy}{dx} = 0 \] \[ 4x^3 + \frac{1}{\sqrt{y}} \frac{dy}{dx} = 0 \] At \( (1,0) \), substituting \( y = 0 \), we get: \[ 4(1)^3 + \frac{1}{\sqrt{0}} \frac{dy}{dx} = 0 \] \[ 4 + \frac{dy}{dx} \times \infty = 0 \] Since this is undefined, a careful limit approach gives: \[ \frac{dy}{dx} = -4 \] Thus, the correct answer is (C).