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

Find all first partial derivatives. $z=\mathrm{ln}\left({x}^{2}+{y}^{2}+1\right)$
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Step 1
Given
The equation is $z=\mathrm{ln}\left({x}^{2}+{y}^{2}+1\right)$.
Step 2
To find all first partial derivatives .
To find $\frac{dz}{dx}$.
$\frac{dz}{dx}\left(\mathrm{ln}\left({x}^{2}+{y}^{2}+1\right)\right)$
$=\frac{1}{{x}^{2}+{y}^{2}+1}\frac{d}{dx}\left({x}^{+}{y}^{2}+1\right)$
$=\frac{1}{{x}^{2}+{y}^{2}+1}\left(2x\right)$
$\frac{dz}{dx}=\frac{2x}{{x}^{2}+{y}^{2}+1}$
To find $\frac{dz}{dy}$.
$\frac{dz}{dy}=\frac{d}{dy}\left(\mathrm{ln}\left({x}^{2}+{y}^{2}+1\right)\right)$
$=\frac{1}{{x}^{2}+{y}^{2}+1}\frac{d}{dy}\left({x}^{2}+{y}^{2}+1\right)$
$=\frac{1}{{x}^{2}+{y}^{2}+1}\left(2y\right)$
$\frac{dz}{dy}=\frac{2y}{{x}^{2}+{y}^{2}+1}$
Therefore , The all partial derivatives are $\frac{dz}{dx}=\frac{2x}{{x}^{2}+{y}^{2}+1}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\frac{dz}{dy}=\frac{2y}{{x}^{2}+{y}^{2}+1}$.
Jeffrey Jordon