/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 43 Find the first partial derivativ... [FREE SOLUTION] | 91影视

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Chapter 13: Problem 43

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

Short Answer

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

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 \(h\) 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: $$\frac{\partial h}{\partial x} = -\sin(x + y + z) \cdot 1 = -\sin(x + y + z)$$
02

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: $$\frac{\partial h}{\partial y} = -\sin(x + y + z) \cdot 1 = -\sin(x + y + z)$$
03

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: $$\frac{\partial h}{\partial x} = -\sin(x+y+z)$$ $$\frac{\partial h}{\partial y} = -\sin(x+y+z)$$ $$\frac{\partial h}{\partial z} = -\sin(x+y+z)$$

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