Problem 35 - Calculus Early Transcendentals

Question

Verify that \(f_{x y}=f_{y x}\) for the following functions. $$f(x, y)=2 x^{3}+3 y^{2}+1$$

Step-by-Step Solution

Answer

Answer: Yes, the second-order mixed partial derivatives of the given function are equal, as we found that \(f_{xy}(x, y) = f_{yx}(x, y) = 0\).

Step 1: Find the first partial derivative with respect to x, \(f_x\)

To find the first partial derivative of \(f(x,y)\) with respect to x, we need to differentiate the function with respect to x while treating y as a constant. So we have: $$f_x(x, y) = \frac{\partial}{\partial x}(2x^3 + 3y^2 + 1)$$ $$f_x(x, y) = 6x^2$$

Step 2: Find the first partial derivative with respect to y, \(f_y\)

Now, we need to find the first partial derivative of \(f(x,y)\) with respect to y, by differentiating the function with respect to y while treating x as a constant. So we have: $$f_y(x, y) = \frac{\partial}{\partial y}(2x^3 + 3y^2 + 1)$$ $$f_y(x, y) = 6y$$

Step 3: Find the mixed partial derivative \(f_{xy}\)

Now we need to find the second-order mixed partial derivative \(f_{xy}\), which is the derivative of \(f_x(x, y)\) (which we found in step 1) with respect to y. So we have: $$f_{xy}(x, y) = \frac{\partial}{\partial y}(6x^2)$$ $$f_{xy}(x, y) = 0$$

Step 4: Find the mixed partial derivative \(f_{yx}\)

Now, we need to find the second-order mixed partial derivative \(f_{yx}\), which is the derivative of \(f_y(x, y)\) (which we found in step 2) with respect to x. So we have: $$f_{yx}(x, y) = \frac{\partial}{\partial x}(6y)$$ $$f_{yx}(x, y) = 0$$

Step 5: Compare \(f_{xy}\) and \(f_{yx}\)

Finally, we need to compare the mixed partial derivatives \(f_{xy}(x, y)\) and \(f_{yx}(x, y)\), which we found in steps 3 and 4, respectively: $$f_{xy}(x, y) = 0$$ $$f_{yx}(x, y) = 0$$ Since \(f_{xy}(x, y) = f_{yx}(x, y)\), the given statement is true and we can conclude that for the given function, \(f(x, y) = 2x^3 + 3y^2 + 1\), the second-order mixed partial derivatives are equal.