022 Sample Final A, Problem 1

Find all first and second partial derivatives of the following function, and demostrate that the mixed second partials are equal for the function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x, y) = \frac{2xy}{x-y}.}

Foundations:
1)Which derivative rules do you have to use for this problem?
2)What is the partial derivative of xy, with respect to x?
1)You have to use the quotient rule, and product rule. The quotient rule says that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial}{\partial x} \left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - g'(x)f(x)}{g(x)^2}} ,   so Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial}{\partial x} \left(\frac{x^2}{x + 1}\right) = \frac{2x(x + 1) - x^2}{(x + 1)^2}} . The product rule says Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial}{\partial x} f(x)g(x) = f'(x)g(x) + g'(x)f(x) } .   This means Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial}{\partial x} x(x + 1) = (x + 1) + x }
2) The partial derivative is y, since we treat anything not involving x as a constant and take the derivative with respect to x. So Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial}{\partial y} xy = x\frac{\partial}{\partial y} y = x.}

Solution:

Step 1:
First, we start by finding the first partial derivatives. So we have to take the partial derivative of f(x, y)with respect to x, and the partial derivative of f(x, y)with respect to y. This gives us the following:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\frac{\partial}{\partial x} f(x, y)} & = & \displaystyle{\frac{\partial}{\partial x} \left( \frac{2xy}{x - y}\right)}\\ &&\\ & = & \displaystyle{\frac{2y(x - y) -2xy}{(x - y)^2}}\\ &&\\ & = & \displaystyle{\frac{-2y^2}{(x - y)^2}} \end{array}}
This gives us the derivative with respect to x. To find the derivative with respect to y, we do the following:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\frac{\partial}{\partial y}f(x, y)} & = & \displaystyle{\frac{\partial}{\partial y}\left(\frac{2xy}{x - y}\right)}\\ &&\\ & = & \displaystyle{\frac{2x(x - y) +2xy}{(x - y)^2}}\\ &&\\ & = & \displaystyle{\frac{2x^2}{(x - y)^2}} \end{array}}

Step 2:
Now we have to find the 4 second derivatives:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\frac{\partial}{\partial x} \frac{\partial f(x, y)}{\partial x}} & = & \displaystyle{\frac{\partial}{\partial x}\left(\frac{-2y^2}{(x - y)^2}\right)}\\ &&\\ & = & \displaystyle{\frac{0 - 2(x - y)(-2y^2)}{(x - y)^4}}\\ &&\\ & = & \displaystyle{\frac{4xy^2 - 4y^3}{(x - y)^4}} \end{array}}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\frac{\partial}{\partial y} \frac{\partial f(x, y)}{\partial x}} & = & \displaystyle{\frac{\partial}{\partial y}\left(\frac{-2y^2}{(x - y)^2}\right)}\\ &&\\ & = & \displaystyle{\frac{-4y(x - y)^2 -4y^2(x - y)}{(x - y)^4}}\\ &&\\ & = & \displaystyle{\frac{-4y(x^2 - 2xy + y^2) - 4xy^2 + 4y^3}{(x - y)^4}}\\ & = & \displaystyle{\frac{4xy^2 - 4x^2y}{(x - y)^4}} \end{array}}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\frac{\partial}{\partial x} \frac{\partial f(x, y)}{\partial y}} & = & \displaystyle{\frac{\partial}{\partial x}\left(\frac{2x^2}{(x - y)^2}\right)}\\ &&\\ & = & \displaystyle{\frac{4x(x - y)^2 -2(x - y)2x^2}{(x - y)^4}}\\ &&\\ & = & \displaystyle{\frac{4x(x^2 - 2xy + y^2) -4x^3+ 4x^2y}{(x - y)^4}}\\ &&\\ & = & \displaystyle{\frac{4xy^2 -4x^2y}{(x - y)^4}} \end{array}}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\frac{\partial}{\partial y} \frac{\partial f(x, y)}{\partial y}} & = & \displaystyle{\frac{\partial}{\partial y}\left(\frac{2x^2}{(x - y)^2}\right)}\\ &&\\ & = & \displaystyle{\frac{0 + 2(x - y)(2x^2)}{(x - y)^4}}\\ &&\\ & = & \displaystyle{\frac{4x^3 - 4x^2y}{(x - y)^4}} \end{array}}

Final Answer:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \displaystyle{\frac{\partial}{\partial x} f(x, y) = \frac{-2y^2}{(x - y)^2} \qquad \frac{\partial}{\partial y} f(x, y) = \frac{2x^2}{(x - y)^2} \qquad}}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \displaystyle{\frac{\partial}{\partial x} \frac{\partial f(x, y)}{\partial x} = \frac{4xy^2 - 4y^3}{(x - y)^4} \qquad \frac{\partial}{\partial y}\frac{\partial f(x, y)}{\partial x} = \frac{4xy^2 - 4x^2y}{(x - y)^4} \qquad \frac{\partial}{\partial x}\frac{\partial f(x, y)}{\partial y} = \frac{4xy^2 -4x^2y}{(x - y)^4} \qquad \frac{\partial}{\partial y}\frac{\partial f(x, y)}{\partial y} = \frac{4x^3 - 4x^2y}{(x - y)^4}} }

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