Find all the second partial derivatives: $f(x, y) = x^5y^8 + 3x^4y$

Neetesh Kumar | December 3, 2024

Calculus Homework Help

This is the solution to Math 1D
Assignment: 14.3 Question Number 24
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Step-by-step solution:

Step 1: Compute First Partial Derivatives

Step 1.1: Partial derivative with respect to $x$:

$f_x(x, y) = \frac{\partial}{\partial x}(x^5y^8 + 3x^4y)$

$f_x(x, y) = 5x^4y^8 + 12x^3y$

Step 1.2: Partial derivative with respect to $y$:

$f_y(x, y) = \frac{\partial}{\partial y}(x^5y^8 + 3x^4y)$

$f_y(x, y) = 8x^5y^7 + 3x^4$

Step 2: Compute Second Partial Derivatives

Step 2.1: Second partial derivative with respect to $x$:

$f_{xx}(x, y) = \frac{\partial}{\partial x}(f_x(x, y))$

$f_{xx}(x, y) = \frac{\partial}{\partial x}(5x^4y^8 + 12x^3y)$

$f_{xx}(x, y) = 20x^3y^8 + 36x^2y$

Step 2.2: Mixed partial derivative $f_{xy}(x, y)$:

$f_{xy}(x, y) = \frac{\partial}{\partial y}(f_x(x, y))$

$f_{xy}(x, y) = \frac{\partial}{\partial y}(5x^4y^8 + 12x^3y)$

$f_{xy}(x, y) = 40x^4y^7 + 12x^3$

Step 2.3: Mixed partial derivative $f_{yx}(x, y)$:

$f_{yx}(x, y) = \frac{\partial}{\partial x}(f_y(x, y))$

$f_{yx}(x, y) = \frac{\partial}{\partial x}(8x^5y^7 + 3x^4)$

$f_{yx}(x, y) = 40x^4y^7 + 12x^3$

Note: $f_{xy} = f_{yx}$ (Clairaut’s Theorem).

Step 2.4: Second partial derivative with respect to $y$:

$f_{yy}(x, y) = \frac{\partial}{\partial y}(f_y(x, y))$

$f_{yy}(x, y) = \frac{\partial}{\partial y}(8x^5y^7 + 3x^4)$

$f_{yy}(x, y) = 56x^5y^6$

Final Answer:

$f_{xx}(x, y) = \boxed{20x^3y^8 + 36x^2y}$

$f_{xy}(x, y) = \boxed{40x^4y^7 + 12x^3}$

$f_{yx}(x, y) = \boxed{40x^4y^7 + 12x^3}$

$f_{yy}(x, y) = \boxed{56x^5y^6}$

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