Question # Find both first partial derivatives. z = \ln(x^{2}+y^{2})

Derivatives
ANSWERED Find both first partial derivatives. $$\displaystyle{z}={\ln{{\left({x}^{{{2}}}+{y}^{{{2}}}\right)}}}$$ 2021-04-12
Step 1
Given function is $$\displaystyle{z}={\ln{{\left({x}^{{{2}}}+{y}^{{{2}}}\right)}}}$$.
Partial derivative means differentiating with respect to one variable keeping other variable as constant.
Finding partial derivative of given function with respect to x, keeping y as constant.
$$\displaystyle{z}_{{{x}}}={\frac{{\partial}}{{\partial{x}}}}{\left[{\ln{{\left({x}^{{{2}}}+{y}^{{{2}}}\right)}}}\right]}$$
$$\displaystyle={\frac{{{1}}}{{{\left({x}^{{{2}}}+{y}^{{{2}}}\right)}}}}\cdot{\frac{{\partial}}{{\partial{z}}}}{\left({x}^{{{2}}}+{y}^{{{2}}}\right)}$$
$$\displaystyle={\frac{{{1}}}{{{\left({x}^{{{2}}}+{y}^{{{2}}}\right)}}}}\cdot{\left({2}{x}+{0}\right)}$$
$$\displaystyle={\frac{{{2}{x}}}{{{x}^{{{2}}}+{y}^{{{2}}}}}}$$
Step 2
Now, finding the partial derivative of given function with respect to y, keeping x as constant.
$$\displaystyle{z}_{{{y}}}={\frac{{\partial}}{{\partial{z}}}}{\left[{\ln{{\left({x}^{{{2}}}+{y}^{{{2}}}\right)}}}\right]}$$
$$\displaystyle={\frac{{{1}}}{{{\left({x}^{{{2}}}+{y}^{{{2}}}\right)}}}}\cdot{\frac{{\partial}}{{\partial{y}}}}{\left({x}^{{{2}}}+{y}^{{{2}}}\right)}$$
$$\displaystyle={\frac{{{1}}}{{{\left({x}^{{{2}}}+{y}^{{{2}}}\right)}}}}\cdot{\left({0}+{2}{y}\right)}$$
$$\displaystyle={\frac{{{2}{y}}}{{{x}^{{{2}}}+{y}^{{{2}}}}}}$$