Mathematics & Numerical Analysis

According to Clairaut's (or Schwarz's) Theorem on mixed partial derivatives:
If a function $z = f(x, y)$ and its partial derivatives $f_x, f_y, f_{xy}, f_{yx}$ are continuous in an open region, then the mixed second-order partial derivatives are identical:

Since $z = 3x^2 y - x \sin xy$ is composed of polynomial and trigonometric functions, it is infinitely differentiable everywhere on $\mathbb{R}^2$.

Calculations for the Jacobian:

• The direct Jacobian of Cartesian coordinates $(x, y)$ with respect to polar coordinates $(r, \theta)$ is:
  $$J = \frac{\partial(x, y)}{\partial(r, \theta)} = \begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\[6pt] \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{vmatrix} = \begin{vmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{vmatrix}$$
  $$J = r\cos^2\theta - (-r\sin^2\theta) = r(\cos^2\theta + \sin^2\theta) = r$$
• By the reciprocal property of Jacobians:
  $$\frac{\partial(r, \theta)}{\partial(x, y)} = \frac{1}{\frac{\partial(x, y)}{\partial(r, \theta)}} = \frac{1}{r}$$

Calculations:

• First derivative: $f'(x) = 3x^2 + 2ax + b$.
• Since relative extrema occur at critical points $x = -3$ and $x = 1$, these are the roots of $f'(x) = 0$:
  • Sum of roots: $(-3) + 1 = -\frac{2a}{3} \implies -2 = -\frac{2a}{3} \implies a = 3$

• Product of roots: $(-3)(1) = \frac{b}{3} \implies -3 = \frac{b}{3} \implies b = -9$

• To find $c$, use the minimum point $(1, -14)$:
  $$f(1) = 1^3 + 3(1)^2 - 9(1) + c = -14$$
  $$1 + 3 - 9 + c = -14 \implies -5 + c = -14 \implies c = -9$$
• Verification at $(-3, 18)$:
  $$f(-3) = (-27) + 3(9) - 9(-3) - 9 = -27 + 27 + 27 - 9 = 18$$