Question

Find both first partial derivatives. z = \ln( x + y )/( x − y)

Derivatives
ANSWERED
asked 2021-02-17
Find both first partial derivatives. \(\displaystyle{z}=\frac{{\ln{{\left({x}+{y}\right)}}}}{{{x}−{y}}}\)

Answers (1)

2021-02-19
Step 1
Given Data:
Function: \(\displaystyle{z}={\ln{{\frac{{{\left({x}+{y}\right)}}}{{{\left({x}-{y}\right)}}}}}}\)
Rewrite the given function.
\(\displaystyle{z}={\ln{{\left({x}+{y}\right)}}}−{\ln{{\left({x}−{y}\right)}}}\)
The partial derivative of the above function with respect to x is,
\(\displaystyle{\frac{{\partial{z}}}{{\partial{x}}}}={\frac{{\partial}}{{\partial{x}}}}{\left({\ln{{\left({x}+{y}\right)}}}-{\ln{{\left({x}-{y}\right)}}}\right)}\)
\(\displaystyle={\frac{{{1}}}{{{x}+{y}}}}-{\frac{{{1}}}{{{x}-{y}}}}\)
\(\displaystyle={\frac{{{x}-{y}-{\left({x}+{y}\right)}}}{{{x}+{y}{\left({x}-{y}\right)}}}}\)
\(\displaystyle={\frac{{{x}-{y}-{x}-{y}}}{{{x}^{{{2}}}-{y}^{{{2}}}}}}\)
\(\displaystyle={\frac{{-{2}{y}}}{{{x}^{{{2}}}-{y}^{{{2}}}}}}\)
Step 2
The partial derivative of the function with respect to y is,
\(\displaystyle{\frac{{\partial{z}}}{{\partial{y}}}}={\frac{{\partial}}{{\partial{y}}}}{\left({\ln{{\left({x}+{y}\right)}}}-{\ln{{\left({x}-{y}\right)}}}\right)}\)
\(\displaystyle={\frac{{{1}}}{{{x}+{y}}}}-{\frac{{{\left(-{1}\right)}}}{{{x}-{y}}}}\)
\(\displaystyle={\frac{{{x}-{y}+{\left({x}+{y}\right)}}}{{{x}+{y}{\left({x}-{y}\right)}}}}\)
\(\displaystyle={\frac{{{x}-{y}+{x}+{y}}}{{{x}^{{{2}}}-{y}^{{{2}}}}}}\)
\(\displaystyle={\frac{{{2}{x}}}{{{x}^{{{2}}}-{y}^{{{2}}}}}}\)
Thus, the partial derivative with respect to x is \(\displaystyle{\frac{{-{2}{y}}}{{{x}^{{{2}}}-{y}^{{{2}}}}}}\) and the partial derivative with respect y is \(\displaystyle{\frac{{{2}{x}}}{{{x}^{{{2}}}-{y}^{{{2}}}}}}\)
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