Problem 17 - Calculus Early Transcendentals

Question

Find the first partial derivatives of the following functions. $$f(w, z)=\frac{w}{w^{2}+z^{2}}$$

Step-by-Step Solution

Answer

Question: Find the first partial derivatives of the function $$f(w, z)=\frac{w}{w^{2}+z^{2}}$$ with respect to both variables, \(w\) and \(z\). Answer: $$\frac{\partial f}{\partial w}=\frac{z^{2}-w^{2}}{(w^{2}+z^{2})^{2}}$$ $$\frac{\partial f}{\partial z}=\frac{2wz}{(w^{2}+z^{2})^{2}}$$

Step 1: Identify the terms to take derivatives of

We have a given function, $$f(w, z)=\frac{w}{w^{2}+z^{2}}$$ and we will find the first partial derivatives with respect to each variable \(w\) and \(z\).

Step 2: Find the partial derivative with respect to w

To find the partial derivative with respect to \(w\), we will treat all other variables (\(z\) in this case) as constants. Using the quotient rule, we have: $$\frac{\partial}{\partial w}\left(\frac{w}{w^{2}+z^{2}}\right)=\frac{(w^{2}+z^{2})\frac{d}{dw}(w)-w\frac{d}{dw}(w^{2}+z^{2})}{(w^{2}+z^{2})^{2}}$$ Now, \(\frac{d}{dw}(w)=1\) and \(\frac{d}{dw}(w^{2}+z^{2})=2w\). Substitute these in the above equation: $$\frac{\partial f}{\partial w}=\frac{(w^{2}+z^{2})(1)-w(2w)}{(w^{2}+z^{2})^{2}}=\frac{w^{2}+z^{2}-2w^{2}}{(w^{2}+z^{2})^{2}}=\frac{z^{2}-w^{2}}{(w^{2}+z^{2})^{2}}$$ So, we have the first partial derivative of \(f(w, z)\) with respect to \(w\): $$\frac{\partial f}{\partial w}=\frac{z^{2}-w^{2}}{(w^{2}+z^{2})^{2}}$$

Step 3: Find the partial derivative with respect to z

Now, let's find the partial derivative with respect to \(z\). We will treat \(w\) as a constant. Using the quotient rule, we have: $$\frac{\partial}{\partial z}\left(\frac{w}{w^{2}+z^{2}}\right)=\frac{w\frac{d}{dz}(w^{2}+z^{2})-0\cdot(w^{2}+z^{2})}{(w^{2}+z^{2})^{2}}$$ Since \(\frac{d}{dz}(w^{2}+z^{2})=2z\), substitute this into the formula: $$\frac{\partial f}{\partial z}=\frac{w(2z)}{(w^{2}+z^{2})^{2}}=\frac{2wz}{(w^{2}+z^{2})^{2}}$$ So, we have the first partial derivative of \(f(w, z)\) with respect to \(z\): $$\frac{\partial f}{\partial z}=\frac{2wz}{(w^{2}+z^{2})^{2}}$$ The first partial derivatives of \(f(w, z)\) are the following: $$\frac{\partial f}{\partial w}=\frac{z^{2}-w^{2}}{(w^{2}+z^{2})^{2}}$$ $$\frac{\partial f}{\partial z}=\frac{2wz}{(w^{2}+z^{2})^{2}}$$