Question

If $z = x^2 y + e^{xy^2}$, then $\left(\frac{\partial^2 z}{\partial x^2} + \frac{\partial^2 z}{\partial x \partial y}\right)$ evaluated at $(1,0)$ is:}

• A. \(0\)
• B. \(1\)
• C. \(2\)
• D. \(-2\)

Hint:

When computing partial derivatives that must be evaluated at a point where one of the coordinates is $0$, scan terms for high powers of that zero-variable early to eliminate tedious product rule computations. Here, since $y=0$, any term retaining an un-differentiated factor of $y$ drops immediately to zero.

Answer 1

The Correct Option is C

Concept: This problem requires partial differentiation of a function of two independent variables $x$ and $y$. - $\frac{\partial z}{\partial x}$ treats $y$ as a constant. - $\frac{\partial^2 z}{\partial x^2} = \frac{\partial}{\partial x}\left(\frac{\partial z}{\partial x}\right)$. - $\frac{\partial^2 z}{\partial x \partial y} = \frac{\partial}{\partial x}\left(\frac{\partial z}{\partial y}\right)$ or equivalently $\frac{\partial}{\partial y}\left(\frac{\partial z}{\partial x}\right)$ due to Clairaut's theorem for smooth functions.

Step 1: Compute the first-order partial derivatives.
The given multivariate function is: \[ z = x^2 y + e^{xy^2} \] Differentiating $z$ partially with respect to $x$ (treating $y$ as fixed): \[ \frac{\partial z}{\partial x} = \frac{\partial}{\partial x}(x^2 y) + \frac{\partial}{\partial x}(e^{xy^2}) = 2xy + y^2 e^{xy^2} \] Differentiating $z$ partially with respect to $y$ (treating $x$ as fixed): \[ \frac{\partial z}{\partial y} = \frac{\partial}{\partial y}(x^2 y) + \frac{\partial}{\partial y}(e^{xy^2}) = x^2 + 2xy e^{xy^2} \]

Step 2: Find the second-order partial derivative $\frac{\partial^2 z{\partial x^2}$.}
Differentiating $\frac{\partial z}{\partial x}$ again with respect to $x$: \[ \frac{\partial^2 z}{\partial x^2} = \frac{\partial}{\partial x} \left( 2xy + y^2 e^{xy^2} \right) \] Evaluating this term by term with respect to $x$: \[ \frac{\partial^2 z}{\partial x^2} = 2y + y^2 \cdot (y^2 e^{xy^2}) = 2y + y^4 e^{xy^2} \]

Step 3: Find the mixed partial derivative $\frac{\partial^2 z{\partial x \partial y}$.}
Differentiating $\frac{\partial z}{\partial x}$ with respect to $y$: \[ \frac{\partial^2 z}{\partial y \partial x} = \frac{\partial}{\partial y} \left( 2xy + y^2 e^{xy^2} \right) \] Applying the product rule to the second term: \[ \frac{\partial^2 z}{\partial y \partial x} = 2x + \left[ 2y \cdot e^{xy^2} + y^2 \cdot (2xy e^{xy^2}) \right] \] \[ \frac{\partial^2 z}{\partial y \partial x} = 2x + 2y e^{xy^2} + 2xy^3 e^{xy^2} \]

Step 4: Evaluate the expressions at the coordinate point $(x, y) = (1, 0)$.
For $\frac{\partial^2 z}{\partial x^2}$: \[ \left. \frac{\partial^2 z}{\partial x^2} \right|_{(1,0)} = 2(0) + (0)^4 e^{(1)(0)^2} = 0 \] For $\frac{\partial^2 z}{\partial y \partial x}$: \[ \left. \frac{\partial^2 z}{\partial y \partial x} \right|_{(1,0)} = 2(1) + 2(0)e^{0} + 2(1)(0)^3 e^{0} = 2 \]

Step 5: Compute the required sum.
Combining the values calculated: \[ \left( \frac{\partial^2 z}{\partial x^2} + \frac{\partial^2 z}{\partial x \partial y} \right) = 0 + 2 = 2 \] This matches Option (C).