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Chapter 21: Problem 7

If \(w=3 x y^{2}+2 y z^{2}\) find all first partial derivatives of \(w\) at the point with coordinates \((1,2,3)\)

Short Answer

Expert verified
Answer: The first partial derivatives at the point (1,2,3) are: \(\frac{\partial w}{\partial x} = 12\), \(\frac{\partial w}{\partial y} = 84\), and \(\frac{\partial w}{\partial z} = 48\).

Step by step solution

01

Write down the function

We are given the function: \(w=3xy^2+2yz^2\).
02

Compute the partial derivative with respect to x

To compute the partial derivative of \(w\) with respect to x, we treat y and z as constants and differentiate w with respect to x: \(\frac{\partial w}{\partial x} = \frac{\partial}{\partial x}(3xy^2+2yz^2) = 3y^2\).
03

Compute the partial derivative with respect to y

To compute the partial derivative of \(w\) with respect to y, we treat x and z as constants and differentiate w with respect to y: \(\frac{\partial w}{\partial y} = \frac{\partial}{\partial y}(3xy^2+2yz^2) = 6xy + 4yz^2\).
04

Compute the partial derivative with respect to z

To compute the partial derivative of \(w\) with respect to z, we treat x and y as constants and differentiate w with respect to z: \(\frac{\partial w}{\partial z} = \frac{\partial}{\partial z}(3xy^2+2yz^2) = 4y^2z\).
05

Evaluate the partial derivatives at the point (1,2,3)

Now we need to evaluate the partial derivatives at the point \((1,2,3)\). Plug in the coordinates into the partial derivatives: \(\frac{\partial w}{\partial x}(1,2,3) = 3(2)^2 = 12\). \(\frac{\partial w}{\partial y}(1,2,3) = 6(1)(2) + 4(2)(3)^2 =12 + 72 = 84\). \(\frac{\partial w}{\partial z}(1,2,3) = 4(2)^2(3) = 48\). So, the first partial derivatives at the point \((1,2,3)\) are: \(\frac{\partial w}{\partial x} = 12\), \(\frac{\partial w}{\partial y} = 84\), and \(\frac{\partial w}{\partial z} = 48\).

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