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Chapter 21: Problem 3

If \(z=9 x+y^{2}\) evaluate \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\) at the point \((4,-2) .\)

Short Answer

Expert verified
Answer: At the point (4, -2), the partial derivatives are \(\frac{\partial z}{\partial x} = 9\) and \(\frac{\partial z}{\partial y} = -4\).

Step by step solution

01

Find the partial derivatives with respect to x and y.

We'll first find the partial derivatives of the function \(z(x, y)\) with respect to \(x\) and \(y\): \(\frac{\partial z}{\partial x} = \frac{\partial}{\partial x}(9x + y^2)\) Since \(y^2\) is treated as a constant when differentiating with respect to \(x\): \(\frac{\partial z}{\partial x} = 9\) Next, let's find the partial derivative with respect to \(y\): \(\frac{\partial z}{\partial y} = \frac{\partial}{\partial y}(9x + y^2)\) Since \(9x\) is treated as a constant when differentiating with respect to \(y\): \(\frac{\partial z}{\partial y} = 2y\)
02

Evaluate the partial derivatives at the given point (4, -2).

Now, we will evaluate these partial derivatives at the given point \((x, y) = (4, -2)\): \(\frac{\partial z}{\partial x} |_{(4, -2)} = 9\) \(\frac{\partial z}{\partial y} |_{(4, -2)} = 2(-2) = -4\) So the partial derivatives at this point \((4, -2)\) are: \(\frac{\partial z}{\partial x} = 9\) \(\frac{\partial z}{\partial y} = -4\)

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