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Chapter 12: Problem 47

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

Short Answer

Expert verified
Question: Find the first partial derivatives of the function h(x, y, z) = cos(x + y + z). Answer: The first partial derivatives of the function h(x, y, z) = cos(x + y + z) are: $$\frac{\partial h}{\partial x} = -\sin (x+y+z)$$ $$\frac{\partial h}{\partial y} = -\sin (x+y+z)$$ $$\frac{\partial h}{\partial z} = -\sin (x+y+z)$$

Step by step solution

01

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 only the function concerning x. We will use the chain rule. $$\frac{\partial h(x, y, z)}{\partial x} = \frac{\partial(\cos (x+y+z))}{\partial x}$$ Now applying the chain rule: $$\frac{\partial h(x, y, z)}{\partial x} = -\sin (x+y+z)\cdot\frac{\partial (x+y+z)}{\partial x} = -\sin (x+y+z) \cdot 1 = -\sin (x+y+z)$$
02

Partial Derivative with respect to y

To find the partial derivative of h with respect to y, we will treat x and z as constants and differentiate only the function concerning y. We will use the chain rule. $$\frac{\partial h(x, y, z)}{\partial y} = \frac{\partial(\cos (x+y+z))}{\partial y}$$ Now applying the chain rule: $$\frac{\partial h(x, y, z)}{\partial y} = -\sin (x+y+z)\cdot\frac{\partial (x+y+z)}{\partial y} = -\sin (x+y+z) \cdot 1 = -\sin (x+y+z)$$
03

Partial Derivative with respect to z

To find the partial derivative of h with respect to z, we will treat x and y as constants and differentiate only the function concerning z. We will use the chain rule. $$\frac{\partial h(x, y, z)}{\partial z} = \frac{\partial(\cos (x+y+z))}{\partial z}$$ Now applying the chain rule: $$\frac{\partial h(x, y, z)}{\partial z} = -\sin (x+y+z)\cdot\frac{\partial (x+y+z)}{\partial z} = -\sin (x+y+z) \cdot 1 = -\sin (x+y+z)$$ So, the first partial derivatives for the function h(x, y, z) = cos(x + y + z) are: $$\frac{\partial h}{\partial x} = -\sin (x+y+z)$$ $$\frac{\partial h}{\partial y} = -\sin (x+y+z)$$ $$\frac{\partial h}{\partial z} = -\sin (x+y+z)$$

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