Question

Let \(y = \sqrt[p]{x^3y}\). If \(\frac{dy}{dx} = \frac{3}{2}\) when \(y = 1\), then the value of \(p\) is equal to \(\ldots\)

• A. \(1\)
• B. \(2\)
• C. \(3\)
• D. \(4\)

Hint:

Raise both sides to the p-th power to get y^(p-1) = x^3, then differentiate.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The relation \(y = (x^3y)^{1/p}\) can be cleared of the root by raising to the power \(p\).

Step 2: Key Formula or Approach:
\[ y^p = x^3 y \Rightarrow y^{p-1} = x^3 \quad (y \ne 0) \]
Differentiate with respect to \(x\) using the chain rule.

Step 3: Detailed Explanation:
\[ (p-1)y^{p-2}\frac{dy}{dx} = 3x^2 \]
When \(y = 1\), the relation \(y^{p-1} = x^3\) gives \(x^3 = 1\), so \(x = 1\).
Substitute \(y = 1\), \(x = 1\) and \(\dfrac{dy}{dx} = \dfrac32\):
\[ (p-1)\cdot 1\cdot\frac{3}{2} = 3 \Rightarrow p - 1 = 2 \Rightarrow p = 3 \]
Check with \(p = 3\): \(y^2 = x^3\), so \(2y y' = 3x^2\) and at \((1,1)\), \(y' = \tfrac32\). It holds.
Option (A) \(p=1\) would make the relation \(1 = x^3\), which defines no curve. Options (B) and (D) give slopes 3 and 1 at \((1,1)\).

Final Answer:
\(p = 3\), option (C). \[ \boxed{3 \text{ (C)}} \]