Question

Find both first partial derivatives. \(f(x, y)=\sqrt{x^{2}+y^{2}}\)

Short answer

The first partial derivatives of the function are \(\frac{\partial f}{\partial x} = \frac{x}{\sqrt{x^{2}+y^{2}}}\) and \(\frac{\partial f}{\partial y} = \frac{y}{\sqrt{x^{2}+y^{2}}}\).

Step-by-step solution

Finding the partial derivative with respect to x

The partial derivative of \(f\) with respect to \(x\) denoted by \(\frac{\partial f}{\partial x}\) is obtained by differentiating \(f\) with respect to \(x\) while treating \(y\) as a constant. Applying this to the function we get \(\frac{\partial f}{\partial x} = \frac{1}{2}(x^{2}+y^{2})^{-1/2} * 2x = \frac{x}{\sqrt{x^{2}+y^{2}}}\).

Finding the partial derivative with respect to y

Similarly, the partial derivative of \(f\) with respect to \(y\) denoted by \(\frac{\partial f}{\partial y}\) is obtained by differentiating \(f\) with respect to \(y\) while treating \(x\) as a constant. Applying this to the function we get \(\frac{\partial f}{\partial y} = \frac{1}{2}(x^{2}+y^{2})^{-1/2} * 2y = \frac{y}{\sqrt{x^{2}+y^{2}}}\).