Question

If $y = 4\sqrt{x}$, then $\frac{d^2y}{dx^2} =$}

• A. $\frac{8}{y^2} \frac{dy}{dx}$
• B. $\frac{-4}{y^2} \frac{dy}{dx}$
• C. $\frac{-8}{y^2} \frac{dy}{dx}$
• D. $\frac{-2}{y^2} \frac{dy}{dx}$

Hint:

For questions with options relating derivatives, it is often simpler to evaluate the derivative and each option separately rather than trying to transform one expression into another from scratch.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We need to calculate the second derivative of \(y\) with respect to \(x\) and express it in terms of \(y\) and the first derivative \(\frac{dy}{dx}\).
Step 2: Key Formula or Approach:
Calculate \(\frac{dy}{dx}\) and \(\frac{d^2y}{dx^2}\).
The options are all of the form \(\frac{C}{y^2} \frac{dy}{dx}\). Set up the equation \(\frac{d^2y}{dx^2} = \frac{C}{y^2} \frac{dy}{dx}\) and solve for the constant \(C\).
Step 3: Detailed Explanation:
Given \(y = 4x^{1/2}\).
First derivative:
\[ \frac{dy}{dx} = 4 \cdot \frac{1}{2}x^{-1/2} = 2x^{-1/2} = \frac{2}{\sqrt{x}} \] Second derivative:
\[ \frac{d^2y}{dx^2} = 2 \cdot \left(-\frac{1}{2}\right)x^{-3/2} = -x^{-3/2} = -\frac{1}{x\sqrt{x}} \] We need to express \(-\frac{1}{x\sqrt{x}}\) in the form \(\frac{C}{y^2} \frac{dy}{dx}\).
First, express \(y^2\) in terms of \(x\):
\[ y^2 = (4\sqrt{x})^2 = 16x \] Now, construct the term \(\frac{1}{y^2} \frac{dy}{dx}\):
\[ \frac{1}{y^2} \frac{dy}{dx} = \frac{1}{16x} \cdot \frac{2}{\sqrt{x}} = \frac{2}{16x\sqrt{x}} = \frac{1}{8x\sqrt{x}} \] We know that \(\frac{d^2y}{dx^2} = -\frac{1}{x\sqrt{x}}\).
By comparison, multiplying our constructed term by -8 gives the second derivative:
\[ -8 \times \left( \frac{1}{y^2} \frac{dy}{dx} \right) = -8 \times \left( \frac{1}{8x\sqrt{x}} \right) = -\frac{1}{x\sqrt{x}} = \frac{d^2y}{dx^2} \] Thus, \(\frac{d^2y}{dx^2} = \frac{-8}{y^2} \frac{dy}{dx}\).
Step 4: Final Answer:
The relation is \(\frac{-8}{y^2} \frac{dy}{dx}\).