91Ó°ÊÓ

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 \(z\).

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 \(w\): $$\frac{\partial f}{\partial w}=\frac{z^{2}-w^{2}}{(w^{2}+z^{2})^{2}}$$

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: $$\frac{\partial f}{\partial w}=\frac{z^{2}-w^{2}}{(w^{2}+z^{2})^{2}}$$ $$\frac{\partial f}{\partial z}=\frac{2wz}{(w^{2}+z^{2})^{2}}$$