Example Find the partial derivatives $f_{x}, f_{y}$ of the following function.

$$f(x,y) = \sqrt{x^2 + y^2}$$

Answer The partial derivative with respect to $x$ written $f_{x}$. To find $f_{x}$, we treat all variables except $x$ as constants. Then differentiate with respect to $x$．Similarly, the partial derivative with respect to $y$ is given by $f_{y}$ and it can be found by treating all variables except $y$ as constants. Then differentiate with respect to $y$.

$$f_{x}(x,y) = \frac{1}{2\sqrt{x^2 + y^2}}\frac{\partial (x^2+y^2)}{\partial x}$$

$$= \frac{1}{2\sqrt{x^2+y^2}}(2x) = \frac{x}{\sqrt{x^2 + y^2}}$$

$$f_{y}(x,y) = \frac{1}{2\sqrt{x^2 + y^2}}\frac{\partial (x^2+y^2)}{\partial y}$$

$$= \frac{1}{2\sqrt{x^2+y^2}}(2y) = \frac{y}{\sqrt{x^2 + y^2}}$$