Problem 37 - Calculus Early Transcendentals

Question

Verify that \(f_{x y}=f_{y x}\) for the following functions. $$f(x, y)=\cos x y$$

Step-by-Step Solution

Answer

Question: Verify whether the second partial derivatives with respect to x and y are equal for the function \(f(x, y) = \cos(xy)\). Answer: The second partial derivatives, \(f_{xy}(x, y)\) and \(f_{yx}(x, y)\), are not equal for the function \(f(x, y) = \cos(xy)\).

Step 1: Find the first partial derivatives with respect to x and y

To find the partial derivatives, we'll first need to find the first derivatives with respect to x and y: $$f_x(x, y) = \frac{\partial}{\partial x}(\cos(xy))$$ $$f_y(x, y) = \frac{\partial}{\partial y}(\cos(xy))$$

Step 2: Calculate the first derivatives

Applying the chain rule, we get: $$f_x(x, y) = -\sin(xy) \cdot y$$ $$f_y(x, y) = -\sin(xy) \cdot x$$

Step 3: Calculate the second partial derivatives

Now, we need to compute the second partial derivatives. We will find the partial derivative of \(f_x(x, y)\) with respect to y and the partial derivative of \(f_y(x, y)\) with respect to x: $$f_{xy}(x, y) = \frac{\partial}{\partial y}(-\sin(xy) \cdot y)$$ $$f_{yx}(x, y) = \frac{\partial}{\partial x}(-\sin(xy) \cdot x)$$

Step 4: Calculate the second derivatives

Again, by applying the chain rule, we obtain: $$f_{xy}(x, y) = -\cos(xy) \cdot y^2 - \sin(xy) \cdot x$$ $$f_{yx}(x, y) = -\cos(xy) \cdot x^2 - \sin(xy) \cdot y$$

Step 5: Compare the second derivatives

Now, let's compare the results for \(f_{xy}(x, y)\) and \(f_{yx}(x, y)\): $$f_{xy}(x, y) = -\cos(xy) \cdot y^2 - \sin(xy) \cdot x$$ $$f_{yx}(x, y) = -\cos(xy) \cdot x^2 - \sin(xy) \cdot y$$ Since the second derivatives \(f_{xy}(x, y)\) and \(f_{yx}(x, y)\) are not equal, we have shown that for this function, \(f_{xy}\) does not equal \(f_{yx}\).