Question

Use the limit definition of partial derivatives to find \(f_{x}(x, y)\) and \(f_{y}(x, y)\). \(f(x, y)=\frac{1}{x+y}\)

Short answer

The partial derivative of the function with respect to \(x\) is \(f_x(x, y) = -\frac{1}{(x+y)^2}\) and with respect to \(y\) is \(f_y(x, y) = -\frac{1}{(x+y)^2}\).

Step-by-step solution

Formulate the Definition of Partial Derivatives

The limit definition of a partial derivative with respect to \(x\) is given by: \(f_x(x, y) = \lim_{h \to 0} \dfrac{f(x+h, y) - f(x, y)}{h}\). For the partial derivative with respect to \(y\), it's given by: \(f_y(x, y) = \lim_{k \to 0} \dfrac{f(x, y+k) - f(x, y)}{k}\) These definitions will be used to find the partial derivatives.

Compute \(f_x(x, y)\)

Substituting \(f(x,y)\) into the definition for \(f_x(x, y)\), gives: \[f_x(x, y) = \lim_{h \to 0} \frac{1}{(x+h)+y} - \frac{1}{(x+y)} \div h = \lim_{h \to 0} \frac{1}{h} \cdot \frac{y-x-h}{(x+y)(x+y+h)} = -\frac{1}{(x+y)^2}\] after simplification.

Compute \(f_y(x, y)\)

Using a similar process, we substitute \(f(x,y)\) into the definition for \(f_y(x, y)\) and get: \[f_y(x, y) = \lim_{k \to 0} \frac{1}{x+(y+k)} - \frac{1}{(x+y)} \div k = \lim_{k \to 0} \frac{1}{k} \cdot \frac{x-y-k}{(x+y)(x+y+k)} = -\frac{1}{(x+y)^2}\] after simplification. The limit computation involved rationalizing the denominator in both cases.