Problem 26 - Calculus Early Transcendentals

Question

Find the four second partial derivatives of the following functions. $$f(x, y)=2 x^{5} y^{2}+x^{2} y$$

Step-by-Step Solution

Answer

Answer: The four second partial derivatives are: $$f_{xx}(x, y) = 40x^3y^2+2y$$ $$f_{yy}(x, y) = 4x^5$$ $$f_{xy}(x, y) = 20x^4y+2x$$ $$f_{yx}(x, y) = 20x^4y+2x$$

Step 1: Find the first partial derivatives

First, we will find the two first partial derivatives of the function with respect to x and y, respectively. $$f_x(x, y)=\frac{\partial}{\partial x}(2x^5y^2+x^2y)=10x^4y^2+2xy$$ $$f_y(x, y)=\frac{\partial}{\partial y}(2x^5y^2+x^2y)=4x^5y+ x^2$$

Step 2: Find the second partial derivatives

Now, we will find the four second partial derivatives by taking the derivative of the first partial derivatives that we just found: 1) $$f_{xx}(x, y) = \frac{\partial}{\partial x} (10x^4y^2+2xy) =40x^3y^2 + 2y$$ 2) $$f_{yy}(x, y) = \frac{\partial}{\partial y} (4x^5y+ x^2) = 4x^5$$ 3) $$f_{xy}(x, y) = \frac{\partial}{\partial y} (10x^4y^2+2xy)=20x^4y+2x$$ 4) $$f_{yx}(x, y) = \frac{\partial}{\partial x} (4x^5y+ x^2) = 20x^4y+2x$$ Note that $$f_{xy}(x, y) = f_{yx}(x, y)$$. This is a general property of differentiable functions. So, the four second partial derivatives are: $$f_{xx}(x, y) = 40x^3y^2+2y$$ $$f_{yy}(x, y) = 4x^5$$ $$f_{xy}(x, y) = 20x^4y+2x$$ $$f_{yx}(x, y) = 20x^4y+2x$$