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Chapter 13: Problem 49

Find the first partial derivatives of the following functions. $$h(w, x, y, z)=\frac{w z}{x y}$$

Short Answer

Expert verified
Answer: The first partial derivatives of the function h(w, x, y, z) are: $$\frac{\partial h}{\partial w} = \frac{z}{x y}, \ \frac{\partial h}{\partial x} = \frac{-w z}{x^2 y}, \ \frac{\partial h}{\partial y} = \frac{-w z}{x y^2}, \ \frac{\partial h}{\partial z} = \frac{w}{x y}$$

Step by step solution

01

Partial derivative with respect to w

To find the partial derivative of the function with respect to w, we differentiate h(w, x, y, z) with respect to w while treating x, y, and z as constants. Using the power rule, we get: $$\frac{\partial h}{\partial w} = \frac{\partial}{\partial w}\left(\frac{w z}{x y}\right) = \frac{z}{x y}$$
02

Partial derivative with respect to x

Now let's find the partial derivative of the function with respect to x. We differentiate h(w, x, y, z) with respect to x while treating w, y, and z as constants. This time, we'll use the quotient rule: $$\frac{\partial h}{\partial x} = \frac{\partial}{\partial x}\left(\frac{w z}{x y}\right) = \frac{-w z}{x^2 y}$$
03

Partial derivative with respect to y

Now we will find the partial derivative of the function with respect to y. Similarly, differentiate h(w, x, y, z) with respect to y while treating w, x, and z as constants. Using the quotient rule, we obtain: $$\frac{\partial h}{\partial y} = \frac{\partial}{\partial y}\left(\frac{w z}{x y}\right) = \frac{-w z}{x y^2}$$
04

Partial derivative with respect to z

Finally, let's find the partial derivative of the function with respect to z. Differentiate h(w, x, y, z) with respect to z while treating w, x, and y as constants. Using the power rule, we get: $$\frac{\partial h}{\partial z} = \frac{\partial}{\partial z}\left(\frac{w z}{x y}\right) = \frac{w}{x y}$$ In summary, the first partial derivatives of the function h(w, x, y, z) are: $$\frac{\partial h}{\partial w} = \frac{z}{x y}, \ \frac{\partial h}{\partial x} = \frac{-w z}{x^2 y}, \ \frac{\partial h}{\partial y} = \frac{-w z}{x y^2}, \ \frac{\partial h}{\partial z} = \frac{w}{x y}$$

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Most popular questions from this chapter

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