Given \( f(x, y)=8 x^{6} \cos \left(y^{5}\right) \), find \( f_{x y}(x, y)= \) \( f_{y y}(x, y)= \)

To find the mixed partial derivative \( f_{xy}(x, y) \) and the second derivative \( f_{yy}(x, y) \) for the function \( f(x, y) = 8 x^6 \cos(y^5) \), we will first find the first partial derivatives. 1. **Finding \( f_x(x, y) \)**: \[ f_x(x, y) = \frac{\partial}{\partial x}(8 x^6 \cos(y^5)) = 48 x^5 \cos(y^5) \] 2. **Finding \( f_y(x, y) \)**: \[ f_y(x, y) = \frac{\partial}{\partial y}(8 x^6 \cos(y^5)) = 8 x^6 \cdot \frac{\partial}{\partial y}(\cos(y^5)) = 8 x^6 \cdot (-\sin(y^5) \cdot 5y^4) = -40 x^6 y^4 \sin(y^5) \] 3. **Finding \( f_{xy}(x, y) \)**: Now, we take the partial derivative of \( f_x \) with respect to \( y \): \[ f_{xy}(x, y) = \frac{\partial}{\partial y}(48 x^5 \cos(y^5)) = 48 x^5 \cdot \frac{\partial}{\partial y}(\cos(y^5)) = 48 x^5 \cdot (-\sin(y^5) \cdot 5y^4) = -240 x^5 y^4 \sin(y^5) \] 4. **Finding \( f_{yy}(x, y) \)**: Now we take the partial derivative of \( f_y \) with respect to \( y \): \[ f_{yy}(x, y) = \frac{\partial}{\partial y}(-40 x^6 y^4 \sin(y^5)) = -40 x^6 \cdot \left(4y^3 \sin(y^5) + y^4 \cdot \frac{\partial}{\partial y}(\sin(y^5))\right) \] Using the chain rule for \( \sin(y^5) \): \[ \frac{\partial}{\partial y}(\sin(y^5)) = \cos(y^5) \cdot 5y^4 \] Thus, \[ f_{yy}(x, y) = -40 x^6 \left(4y^3 \sin(y^5) + y^4 \cdot (5y^4 \cos(y^5))\right) \] Simplifying: \[ f_{yy}(x, y) = -40 x^6 \left(4y^3 \sin(y^5) + 5y^8 \cos(y^5)\right) \] In conclusion, we have: \[ f_{xy}(x, y) = -240 x^5 y^4 \sin(y^5) \] \[ f_{yy}(x, y) = -40 x^6 \left(4y^3 \sin(y^5) + 5y^8 \cos(y^5)\right) \]