Question: Find the four second partial derivatives of the following function.
f(x,y) = y^4 cos 6x
1. **State the problem:** We need to find the four second partial derivatives of the function $$f(x,y) = y^4 \cos 6x$$.
2. **Recall the definitions:** The four second partial derivatives are:
- $$f_{xx} = \frac{\partial^2 f}{\partial x^2}$$
- $$f_{yy} = \frac{\partial^2 f}{\partial y^2}$$
- $$f_{xy} = \frac{\partial^2 f}{\partial x \partial y}$$
- $$f_{yx} = \frac{\partial^2 f}{\partial y \partial x}$$
3. **First, find the first partial derivatives:**
- $$f_x = \frac{\partial}{\partial x} (y^4 \cos 6x) = y^4 \frac{\partial}{\partial x} (\cos 6x) = y^4 (-6 \sin 6x) = -6 y^4 \sin 6x$$
- $$f_y = \frac{\partial}{\partial y} (y^4 \cos 6x) = \cos 6x \frac{\partial}{\partial y} (y^4) = \cos 6x (4 y^3) = 4 y^3 \cos 6x$$
4. **Now, find the second partial derivatives:**
- $$f_{xx} = \frac{\partial}{\partial x} (f_x) = \frac{\partial}{\partial x} (-6 y^4 \sin 6x) = -6 y^4 \frac{\partial}{\partial x} (\sin 6x) = -6 y^4 (6 \cos 6x) = -36 y^4 \cos 6x$$
- $$f_{yy} = \frac{\partial}{\partial y} (f_y) = \frac{\partial}{\partial y} (4 y^3 \cos 6x) = \cos 6x \frac{\partial}{\partial y} (4 y^3) = \cos 6x (12 y^2) = 12 y^2 \cos 6x$$
- $$f_{xy} = \frac{\partial}{\partial y} (f_x) = \frac{\partial}{\partial y} (-6 y^4 \sin 6x) = -6 \sin 6x \frac{\partial}{\partial y} (y^4) = -6 \sin 6x (4 y^3) = -24 y^3 \sin 6x$$
- $$f_{yx} = \frac{\partial}{\partial x} (f_y) = \frac{\partial}{\partial x} (4 y^3 \cos 6x) = 4 y^3 \frac{\partial}{\partial x} (\cos 6x) = 4 y^3 (-6 \sin 6x) = -24 y^3 \sin 6x$$
5. **Note:** The mixed partial derivatives $$f_{xy}$$ and $$f_{yx}$$ are equal, which is consistent with Clairaut's theorem.
**Final answers:**
- $$f_{xx} = -36 y^4 \cos 6x$$
- $$f_{yy} = 12 y^2 \cos 6x$$
- $$f_{xy} = f_{yx} = -24 y^3 \sin 6x$$