Question

Evaluate fxy at a point (x,y) for the function f defined by f(x,y)=x (1/tan) y .
Using Schwarz's Theorem evaluate fyx at the point (x,y) .

Answer 1

Answer on Question#41419, Math, Integral Calculus

Evaluate fxy at a point $(x,y)$ for the function $f$ defined by $f(x,y)=x$ (1/tan y). Using Schwarz's Theorem evaluate fyx at the point $(x,y)$.

Solution.

$$f(x, y) = \frac{x}{\tan y}$$ $$f_x(x, y) = \frac{1}{\tan y}$$ $$f_{xy}(x, y) = \left(\frac{1}{\tan y}\right)_y = -\frac{1}{\tan^2 y} \left(\frac{1}{\cos^2 y}\right) = -\frac{1}{\sin^2 y}$$

THEOREM (H. A. Schwarz). Suppose that $f$ is a function of two variables such that $f_{xy}''$ and $f_{yx}''$ both exist and are continuous at some point $(x_0; y_0)$. Then

$$f_{xy}''(x_0; y_0) = f_{yx}''(x_0; y_0)$$

Thus,

$$f_{yx}(x, y) = f_{xy}(x, y) = -\frac{1}{\sin^2 y} = -\csc y$$

Answer: $f_{yx}(x, y) = f_{xy}(x, y) = -\csc y$

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