Question

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

Derivatives
ANSWERED
asked 2021-02-14
Find both first partial derivatives. \(\displaystyle{z}={\frac{{{\ln{{x}}}}}{{{y}}}}\)

Answers (1)

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 \(\displaystyle{z}={\frac{{{\ln{{x}}}}}{{{y}}}}{i}{s}{\frac{{{p}{a}{r}{t}{i}{c}{a}{l}{z}}}{{{p}{a}{r}{t}{i}{c}{a}{l}{x}}}}={\frac{{{1}}}{{{x}{y}}}}{\quad\text{and}\quad}{\frac{{{p}{a}{r}{t}{i}{c}{a}{l}{z}}}{{{p}{a}{r}{t}{i}{c}{a}{l}{y}}}}=-{\frac{{{\ln{{x}}}}}{{{y}^{{{2}}}}}}\)
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