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Chapter 13: Problem 13

Find the first partial derivatives of the following functions. $$g(x, y)=\cos 2 x y$$

Short Answer

Expert verified
The first partial derivatives of the function \(g(x, y) = \cos(2xy)\) are: 1. With respect to x: \(\frac{\partial g(x, y)}{\partial x} = -2y\sin(2xy)\) 2. With respect to y: \(\frac{\partial g(x, y)}{\partial y} = -2x\sin(2xy)\)

Step by step solution

01

Differentiate g(x, y) with respect to x

To find the partial derivative of \(g(x, y)\) with respect to \(x\), we'll keep \(y\) constant and differentiate \(g(x,y)\) with respect to \(x\). Using the chain rule, we find: $$\frac{\partial g(x,y)}{\partial x} = \frac{d (\cos(2xy))}{d (2xy)} \cdot \frac{\partial (2xy)}{\partial x}$$
02

Calculate the derivatives and find the partial derivative with respect to x

Now we calculate the respective derivatives: $$\frac{d (\cos(2xy))}{d (2xy)} = -\sin(2xy)$$ and $$\frac{\partial (2xy)}{\partial x} = 2y$$ Multiplying these derivatives together, we find the partial derivative with respect to \(x\): $$\frac{\partial g(x, y)}{\partial x} = -2y\sin(2xy)$$
03

Differentiate g(x, y) with respect to y

To find the partial derivative of \(g(x, y)\) with respect to \(y\), we'll keep \(x\) constant and differentiate \(g(x, y)\) with respect to \(y\). Using the chain rule, we find: $$\frac{\partial g(x, y)}{\partial y} = \frac{d(\cos(2xy))}{d(2xy)} \cdot \frac{\partial(2xy)}{\partial y}$$
04

Calculate the derivatives and find the partial derivative with respect to y

Now we calculate the respective derivatives: $$\frac{d (\cos(2xy))}{d (2xy)} = -\sin(2xy)$$ and $$\frac{\partial (2xy)}{\partial y} = 2x$$ Multiplying these derivatives together, we find the partial derivative with respect to \(y\): $$\frac{\partial g(x, y)}{\partial y} = -2x\sin(2xy)$$
05

Write down the first partial derivatives

The first partial derivatives of the function \(g(x, y)=\cos(2xy)\) are: $$\frac{\partial g(x, y)}{\partial x} = -2y\sin(2xy)$$ and $$\frac{\partial g(x, y)}{\partial y} = -2x\sin(2xy)$$

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