Question

Find the first partial derivatives with respect to \(x, y\), and \(z\). $$ w=\sqrt{x^{2}+y^{2}+z^{2}} $$

Short answer

The first order partial derivatives of \(w\) are \(\frac{\partial w}{\partial x} = \frac{x}{\sqrt{x^2 + y^2 + z^2}}\), \(\frac{\partial w}{\partial y} = \frac{y}{\sqrt{x^2 + y^2 + z^2}}\) , and \(\frac{\partial w}{\partial z} = \frac{z}{\sqrt{x^2 + y^2 + z^2}}\).

Step-by-step solution

Recognize the structure

The initial step is recognizing the structure of the function \(w = \sqrt{x^2 + y^2 + z^2}\). This function can be written as \(w = (x^2 + y^2 + z^2)^{1/2}\), which is easier for differentiation.

Partially Differentiate with respect to \(x\)

By using the chain rule for differentiation, the derivative of \(w\) with respect to \(x\) is \(\frac{\partial w}{\partial x} = \frac{1}{2}(x^2 + y^2 + z^2)^{-1/2} * 2x = \frac{x}{\sqrt{x^2 + y^2 + z^2}} \).

Partially Differentiate with respect to \(y\)

Following the same procedure as in Step 2, differentiating \(w\) with respect to \(y\) gives \(\frac{\partial w}{\partial y} = \frac{y}{\sqrt{x^2 + y^2 + z^2}}\).

Partially differentiate with respect to \(z\)

Again the same approach gives \(\frac{\partial w}{\partial z} = \frac{z}{\sqrt{x^2 + y^2 + z^2}}\).