Question

If $z = x^4 \sin(xy^3)$, then the value of $\frac{\partial z}{\partial x}$ at the point (0, 0) is given by}

• A. 1
• B. 0
• C. $\frac{\pi}{2}$
• D. -1

Hint:

Since $x^3$ and $x^4$ are factors in both terms of the derivative, any point where $x = 0$ will automatically make the entire partial derivative $\frac{\partial z}{\partial x}$ equal to 0.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The partial derivative $\frac{\partial z}{\partial x}$ is calculated by differentiating the function $z$ with respect to $x$ while treating $y$ as a constant.
Key Formula or Approach:
We use the product rule of differentiation:
\[ \frac{\partial}{\partial x}[u(x) \cdot v(x)] = u'(x) \cdot v(x) + u(x) \cdot v'(x) \]

Step 2: Detailed Explanation:
We are given:
\[ z = x^4 \sin(xy^3) \]
Differentiate with respect to $x$ using the product rule:
\[ \frac{\partial z}{\partial x} = \frac{\partial}{\partial x}(x^4) \cdot \sin(xy^3) + x^4 \cdot \frac{\partial}{\partial x}(\sin(xy^3)) \]
\[ \frac{\partial z}{\partial x} = 4x^3 \sin(xy^3) + x^4 \cos(xy^3) \cdot y^3 \]
Now, substitute the coordinates of the point $(x = 0, y = 0)$ into this derivative expression:
\[ \left.\frac{\partial z}{\partial x}\right|_{(0,0)} = 4(0)^3 \sin(0 \cdot 0^3) + (0)^4 \cos(0 \cdot 0^3) \cdot 0^3 \]
\[ \left.\frac{\partial z}{\partial x}\right|_{(0,0)} = 0 + 0 = 0 \]
Therefore, the value of the partial derivative at $(0, 0)$ is 0.

Step 3: Final Answer
The correct option is (B).