Problem 41
Question
Find the first partial derivatives of the following functions. $$f(x, y, z)=x y+x z+y z$$
Step-by-Step Solution
Verified Answer
Question: Find the first partial derivatives of the function \(f(x, y, z) = xy + xz + yz\).
Answer: The first partial derivatives of the function are as follows:
\(\frac{\partial f}{\partial x} = y + z\)
\(\frac{\partial f}{\partial y} = x + z\)
\(\frac{\partial f}{\partial z} = x + y\)
1Step 1: Find the partial derivative with respect to x
To find the partial derivative with respect to \(x\), we treat \(y\) and \(z\) as constant and differentiate the function \(f(x, y, z)\). Therefore, we will have:
$$\frac{\partial f}{\partial x} = \frac{\partial (x y + x z + y z)}{\partial x}$$
Now, apply the derivative on the function:
$$\frac{\partial f}{\partial x} = y + z$$
2Step 2: Find the partial derivative with respect to y
To find the partial derivative with respect to \(y\), we treat \(x\) and \(z\) as constant and differentiate the function \(f(x, y, z)\). Therefore, we will have:
$$\frac{\partial f}{\partial y} = \frac{\partial (x y + x z + y z)}{\partial y}$$
Now, apply the derivative on the function:
$$\frac{\partial f}{\partial y} = x + z$$
3Step 3: Find the partial derivative with respect to z
To find the partial derivative with respect to \(z\), we treat \(x\) and \(y\) as constant and differentiate the function \(f(x, y, z)\). Therefore, we will have:
$$\frac{\partial f}{\partial z} = \frac{\partial (x y + x z + y z)}{\partial z}$$
Now, apply the derivative on the function:
$$\frac{\partial f}{\partial z} = x + y$$
In conclusion, the first partial derivatives of the function \(f(x, y, z)=xy+xz+yz\) are:
$$\frac{\partial f}{\partial x} = y + z$$
$$\frac{\partial f}{\partial y} = x + z$$
$$\frac{\partial f}{\partial z} = x + y$$
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