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# Find all first partial derivatives. z = ln(x^2 + y^2 + 1)

Question
Derivatives
asked 2021-02-21
Find all first partial derivatives. $$\displaystyle{z}={\ln{{\left({x}^{{2}}+{y}^{{2}}+{1}\right)}}}$$

## Answers (1)

2021-02-22
Step 1
Given
The equation is $$\displaystyle{z}={\ln{{\left({x}^{{2}}+{y}^{{2}}+{1}\right)}}}$$.
Step 2
To find all first partial derivatives .
To find $$\displaystyle\frac{{{\left.{d}{z}\right.}}}{{{\left.{d}{x}\right.}}}$$.
$$\displaystyle\frac{{{\left.{d}{z}\right.}}}{{{\left.{d}{x}\right.}}}{\left({\ln{{\left({x}^{{2}}+{y}^{{2}}+{1}\right)}}}\right)}$$
$$\displaystyle=\frac{{1}}{{{x}^{{2}}+{y}^{{2}}+{1}}}\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({x}^{+}{y}^{{2}}+{1}\right)}$$
$$\displaystyle=\frac{{1}}{{{x}^{{2}}+{y}^{{2}}+{1}}}{\left({2}{x}\right)}$$
$$\displaystyle\frac{{{\left.{d}{z}\right.}}}{{{\left.{d}{x}\right.}}}=\frac{{{2}{x}}}{{{x}^{{2}}+{y}^{{2}}+{1}}}$$
To find $$\displaystyle\frac{{{\left.{d}{z}\right.}}}{{{\left.{d}{y}\right.}}}$$.
$$\displaystyle\frac{{{\left.{d}{z}\right.}}}{{{\left.{d}{y}\right.}}}=\frac{{{d}}}{{{\left.{d}{y}\right.}}}{\left({\ln{{\left({x}^{{2}}+{y}^{{2}}+{1}\right)}}}\right)}$$
$$\displaystyle=\frac{{1}}{{{x}^{{2}}+{y}^{{2}}+{1}}}\frac{{d}}{{{\left.{d}{y}\right.}}}{\left({x}^{{2}}+{y}^{{2}}+{1}\right)}$$
$$\displaystyle=\frac{{1}}{{{x}^{{2}}+{y}^{{2}}+{1}}}{\left({2}{y}\right)}$$
$$\displaystyle\frac{{{\left.{d}{z}\right.}}}{{{\left.{d}{y}\right.}}}=\frac{{{2}{y}}}{{{x}^{{2}}+{y}^{{2}}+{1}}}$$
Therefore , The all partial derivatives are $$\displaystyle\frac{{{\left.{d}{z}\right.}}}{{{\left.{d}{x}\right.}}}=\frac{{{2}{x}}}{{{x}^{{2}}+{y}^{{2}}+{1}}}{\quad\text{and}\quad}\frac{{{\left.{d}{z}\right.}}}{{{\left.{d}{y}\right.}}}=\frac{{{2}{y}}}{{{x}^{{2}}+{y}^{{2}}+{1}}}$$.

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