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

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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: $$\frac{\partial f}{\partial x} = y + z$$ $$\frac{\partial f}{\partial y} = x + z$$ $$\frac{\partial f}{\partial z} = x + y$$