Question

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

Short answer

The partial derivative of the function \(f(x, y)=\sqrt{x+y}\) with respect to \(x\) is \(\frac{1}{2\sqrt{x+y}}\) and with respect to \(y\) is also \(\frac{1}{2\sqrt{x+y}}\).

Step-by-step solution

Write Down the Function

The first step is to write down the given function, which is \(f(x, y)=\sqrt{x+y}\).

Compute the Partial Derivative With Respect to x

Compute the partial derivative of the function with respect to \(x\) while treating \(y\) as a constant. Using the limit definition of the derivative, this gives: \[f_{x}(x, y)= \lim_{h\to 0} \frac{f(x+h, y) - f(x, y)}{h}\]Substituting the function \(f(x, y)\) into the formula and simplifying provides the derivative \(\frac{1}{2\sqrt{x+y}}\).

Compute the Partial Derivative With Respect to y

Now, compute the partial derivative of the function with regards to \(y\), treating \(x\) as a constant. Again, use the limit definition of the derivative: \[f_{y}(x, y)= \lim_{k\to 0} \frac{f(x, y+k) - f(x, y)}{k}\]Substitute the function \(f(x, y)\) into the limit, and then simplify the equation in the same way as done for \(x\), yielding the derivative \(\frac{1}{2\sqrt{x+y}}\).