Question

# Find both first partial derivatives. z=\frac{\ln x}{y}

Derivatives
Find both first partial derivatives. $$\displaystyle{z}={\frac{{{\ln{{x}}}}}{{{y}}}}$$

2021-02-16
Step 1
Given that
$$\displaystyle{z}={\frac{{{\ln{{x}}}}}{{{y}}}}$$
Step 2
On partial differentiating z with respect to x and y,
$$\displaystyle{\frac{{{p}{a}{r}{t}{i}{c}{a}{l}{z}}}{{{p}{a}{r}{t}{i}{c}{a}{l}{x}}}}={\frac{{{p}{a}{r}{t}{i}{c}{a}{l}}}{{{p}{a}{r}{t}{i}{c}{a}{l}{x}}}}{\left({\frac{{{\ln{{x}}}}}{{{y}}}}\right)}$$
$$\displaystyle={\frac{{{1}}}{{{y}}}}{\frac{{{p}{a}{r}{t}{i}{c}{a}{l}}}{{{p}{a}{r}{t}{i}{c}{a}{l}{x}}}}{\left({\ln{{x}}}\right)}$$
$$\displaystyle={\frac{{{1}}}{{{y}}}}{\left({\frac{{{1}}}{{{x}}}}\right)}$$
$$\displaystyle={\frac{{{1}}}{{{x}{y}}}}$$
And
$$\displaystyle{\frac{{{p}{a}{r}{t}{i}{c}{a}{l}{z}}}{{{p}{a}{r}{t}{i}{c}{a}{l}{y}}}}={\frac{{{p}{a}{r}{t}{i}{c}{a}{l}}}{{{p}{a}{r}{t}{i}{c}{a}{l}{y}}}}{\left({\frac{{{\ln{{x}}}}}{{{y}}}}\right)}$$
$$\displaystyle={\ln{{x}}}{\frac{{{p}{a}{r}{t}{i}{c}{a}{l}}}{{{p}{a}{r}{t}{i}{c}{a}{l}{x}}}}{\left({\frac{{{1}}}{{{y}}}}\right)}$$
$$\displaystyle={\ln{{x}}}{\left(-{\frac{{{1}}}{{{y}^{{{2}}}}}}\right)}$$
$$\displaystyle=-{\frac{{{\ln{{x}}}}}{{{y}^{{{2}}}}}}$$
Step 3
Hence first partial derivative of $$\displaystyle{z}={\frac{{{\ln{{x}}}}}{{{y}}}}{i}{s}{\frac{{{p}{a}{r}{t}{i}{c}{a}{l}{z}}}{{{p}{a}{r}{t}{i}{c}{a}{l}{x}}}}={\frac{{{1}}}{{{x}{y}}}}{\quad\text{and}\quad}{\frac{{{p}{a}{r}{t}{i}{c}{a}{l}{z}}}{{{p}{a}{r}{t}{i}{c}{a}{l}{y}}}}=-{\frac{{{\ln{{x}}}}}{{{y}^{{{2}}}}}}$$