Problem 43

Question

Find the first partial derivatives of the following functions. $$h(x, y, z)=\cos (x+y+z)$$

Step-by-Step Solution

Verified
Answer
Answer: The first partial derivatives of the function are: $$\frac{\partial h}{\partial x} = -\sin(x+y+z)$$ $$\frac{\partial h}{\partial y} = -\sin(x+y+z)$$ $$\frac{\partial h}{\partial z} = -\sin(x+y+z)$$
1Step 1: Partial Derivative with respect to x
To find the partial derivative of \(h\) with respect to \(x\), we will treat \(y\) and \(z\) as constants and differentiate \(h\) with respect to \(x\). Using the chain rule, we have: $$\frac{\partial h}{\partial x} = \frac{d \cos(x + y + z)}{d (x + y + z)} \cdot \frac{d(x + y + z)}{d x}$$ Since \(\frac{d \cos(t)}{d t} = -\sin(t)\) and \(\frac{d (x + y + z)}{d x} = 1\), we have: $$\frac{\partial h}{\partial x} = -\sin(x + y + z) \cdot 1 = -\sin(x + y + z)$$
2Step 2: Partial Derivative with respect to y
Next, we find the partial derivative of \(h\) with respect to \(y\) by treating \(x\) and \(z\) as constants and differentiating \(h\) with respect to \(y\). Using the chain rule, we get: $$\frac{\partial h}{\partial y} = \frac{d \cos(x + y + z)}{d (x + y + z)} \cdot \frac{d(x + y + z)}{d y}$$ Since \(\frac{d(x + y + z)}{d y} = 1\), we have: $$\frac{\partial h}{\partial y} = -\sin(x + y + z) \cdot 1 = -\sin(x + y + z)$$
3Step 3: Partial Derivative with respect to z
Finally, we find the partial derivative of \(h\) with respect to \(z\) by treating \(x\) and \(y\) as constants and differentiating \(h\) with respect to \(z\). Using the chain rule, we have: $$\frac{\partial h}{\partial z} = \frac{d \cos(x + y + z)}{d (x + y + z)} \cdot \frac{d(x + y + z)}{d z}$$ Since \(\frac{d (x + y + z)}{d z} = 1\), we have: $$\frac{\partial h}{\partial z} = -\sin(x + y + z) \cdot 1 = -\sin(x + y + z)$$ The first partial derivatives of the function \(h(x, y, z) = \cos(x + y + z)\) are: $$\frac{\partial h}{\partial x} = -\sin(x+y+z)$$ $$\frac{\partial h}{\partial y} = -\sin(x+y+z)$$ $$\frac{\partial h}{\partial z} = -\sin(x+y+z)$$