If 𝑧=𝑠𝑖𝑛⁡(3𝑥+2𝑦), determine 3∂2𝑧∂𝑦2-2∂2𝑧∂𝑥2Question 9Answera.-6𝑠𝑖𝑛⁡(3𝑥+2𝑦)b.12𝑠𝑖𝑛⁡(3𝑥+2𝑦)c.6𝑧d.6𝑐𝑜𝑠⁡(3𝑥+2𝑦)

1. Break Down the Problem

We need to calculate the second partial derivatives of the function $z = \sin(3x + 2y)$ with respect to $y$ and $x$, then evaluate the expression $3\frac{\partial^2 z}{\partial y^2} - 2\frac{\partial^2 z}{\partial x^2}$.

2. Relevant Concepts

• Partial Derivatives: To find $\frac{\partial z}{\partial y}$ and $\frac{\partial z}{\partial x}$, we will use: $\frac{\partial z}{\partial y} = \frac{\partial}{\partial y}(\sin(3x + 2y)) \quad \text{and} \quad \frac{\partial z}{\partial x} = \frac{\partial}{\partial x}(\sin(3x + 2y))$
• Second Partial Derivatives: Then we need to find: $\frac{\partial^2 z}{\partial y^2} \quad \text{and} \quad \frac{\partial^2 z}{\partial x^2}$

3. Analysis and Detail

Step 1: First Partial Derivative with respect to $y$

Using the chain rule: $\frac{\partial z}{\partial y} = \cos(3x + 2y) \cdot \frac{\partial}{\partial y}(3x + 2y) = \cos(3x + 2y) \cdot 2$ So, $\frac{\partial z}{\partial y} = 2\cos(3x + 2y)$

Step 2: Second Partial Derivative with respect to $y$

Now, take the derivative of $\frac{\partial z}{\partial y}$ with respect to $y$: $\frac{\partial^2 z}{\partial y^2} = \frac{\partial}{\partial y}(2\cos(3x + 2y)) = 2(-\sin(3x + 2y) \cdot 2) = -4\sin(3x + 2y)$

Step 3: First Partial Derivative with respect to $x$

Using the chain rule: $\frac{\partial z}{\partial x} = \cos(3x + 2y) \cdot \frac{\partial}{\partial x}(3x + 2y) = \cos(3x + 2y) \cdot 3$ So, $\frac{\partial z}{\partial x} = 3\cos(3x + 2y)$

Step 4: Second Partial Derivative with respect to $x$

Now, take the derivative of $\frac{\partial z}{\partial x}$ with respect to $x$: $\frac{\partial^2 z}{\partial x^2} = \frac{\partial}{\partial x}(3\cos(3x + 2y)) = 3(-\sin(3x + 2y) \cdot 3) = -9\sin(3x + 2y)$

Step 5: Combine the derivatives into the expression

Now, substitute these back into the expression: $3\frac{\partial^2 z}{\partial y^2} - 2\frac{\partial^2 z}{\partial x^2} = 3(-4\sin(3x + 2y)) - 2(-9\sin(3x + 2y))$ This simplifies to: $= -12\sin(3x + 2y) + 18\sin(3x + 2y) = 6\sin(3x + 2y)$

4. Verify and Summarize

The calculations have been checked step-by-step, and the conclusion is consistent.

Final Answer

$6\sin(3x + 2y)$

• Correct option: (c) $6z$ since $z = \sin(3x + 2y)$.