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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 are as follows: 1. With respect to w: $$\frac{\partial h}{\partial w} = \frac{z}{x y}$$ 2. With respect to x: $$\frac{\partial h}{\partial x} = -\frac{w z}{x^2 y}$$ 3. With respect to y: $$\frac{\partial h}{\partial y} = -\frac{w z}{x y^2}$$ 4. With respect to z: $$\frac{\partial h}{\partial z} = \frac{w}{x y}$$

Step by step solution

01

Define the function

Define the function $$h(w, x, y, z)=\frac{w z}{x y}$$ which has four variables: w, x, y, and z.
02

Find the partial derivative with respect to w

To find the partial derivative of h with respect to w, treat x, y, and z as constants and differentiate with respect to w: $$\frac{\partial h}{\partial w}=\frac{\partial}{\partial w}\frac{w z}{x y} = \frac{z}{x y}$$
03

Find the partial derivative with respect to x

To find the partial derivative of h with respect to x, treat w, y, and z as constants and differentiate with respect to x: $$\frac{\partial h}{\partial x}=\frac{\partial}{\partial x}\frac{w z}{x y} = -\frac{w z}{x^2 y}$$
04

Find the partial derivative with respect to y

To find the partial derivative of h with respect to y, treat w, x, and z as constants and differentiate with respect to y: $$\frac{\partial h}{\partial y}=\frac{\partial}{\partial y}\frac{w z}{x y} = -\frac{w z}{x y^2}$$
05

Find the partial derivative with respect to z

To find the partial derivative of h with respect to z, treat w, x, and y as constants and differentiate with respect to z: $$\frac{\partial h}{\partial z}=\frac{\partial}{\partial z}\frac{w z}{x y} = \frac{w}{x y}$$
06

Summary of the first partial derivatives

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

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