The partial derivative of natural logarithm of $x$ square plus $y$ squared with respect to $x$ and also with respect to $y$ can be calculated by using the differentiation formulas.

The partial derivative of natural logarithm of sum of the squares of variables $x$ and $y$ with respect to $x$ is written in mathematics as follows.

$\dfrac{\partial}{\partial x}\,\log_{e}{\big(x^2+y^2\big)}$

In differential calculus, there is a formula for finding the derivative of the natural logarithm of a variable but there is no rule for finding the partial derivative of this type of function. It is a composition of two functions. So, chain rule is only solution to find its partial derivative with respect to $x$.

$=\,\,$ $\dfrac{1}{x^2+y^2} \times \dfrac{\partial}{\partial x}\,\big(x^2+y^2\big)$

According to the sum rule of derivatives, the partial derivative of sum of two functions with respect to $x$ can be calculated by the sum of their partial derivatives.

$=\,\,$ $\dfrac{1}{x^2+y^2} \times \bigg(\dfrac{\partial}{\partial x}\,\big(x^2\big)+\dfrac{\partial}{\partial x}\,\big(y^2\big)\bigg)$

The partial derivative of $x$ squared can be calculated by the power rule of derivatives. In this case, the variable $y$ is considered as a constant. So, the partial derivative of $y$ square is zero as per the derivative rule of a constant.

$=\,\,$ $\dfrac{1}{x^2+y^2} \times \big(2 \times x^{2-1}+0\big)$

The process of the finding the partial differentiation is successfully completed and it is time to simplify the whole mathematical expression.

$=\,\,$ $\dfrac{1}{x^2+y^2} \times \big(2 \times x^1\big)$

$=\,\,$ $\dfrac{1}{x^2+y^2} \times \big(2 \times x\big)$

$=\,\,$ $\dfrac{1}{x^2+y^2} \times \big(2x\big)$

$=\,\,$ $\dfrac{1}{x^2+y^2} \times 2x$

$=\,\,$ $\dfrac{1 \times 2x}{x^2+y^2}$

$=\,\,$ $\dfrac{2x}{x^2+y^2}$

The partial derivative of natural logarithm of $x$ squared plus $y$ squared with respect to $y$ is expressed in the following form.

$\dfrac{\partial}{\partial y}\,\log_{e}{\big(x^2+y^2\big)}$

Use the derivative rule of logarithmic function through chain rule to find the partial derivative of natural logarithm of sum of squares of variables $x$ and $y$ with respect to $y$.

$=\,\,$ $\dfrac{1}{x^2+y^2} \times \dfrac{\partial}{\partial y}\,\big(x^2+y^2\big)$

The partial derivative of sum of two functions can be calculated by the sum of their partial derivatives as per the sum rule of the derivatives.

$=\,\,$ $\dfrac{1}{x^2+y^2} \times \bigg(\dfrac{\partial}{\partial y}\,\big(x^2\big)+\dfrac{\partial}{\partial y}\,\big(y^2\big)\bigg)$

The partial differentiation should be done with respect to $y$. So, the function in terms of $x$ is considered as a constant. According to the derivative rule of a constant, the partial derivative of the function $x$ squared is zero. Similarly, use the power rule of derivatives to find the partial differentiation of the square of variable $y$ with respect to $y$.

$=\,\,$ $\dfrac{1}{x^2+y^2} \times \big(0+2 \times y^{2-1}\big)$

It is time to simplify the mathematical expression for expressing the partial derivative of the given function in simplified form.

$=\,\,$ $\dfrac{1}{x^2+y^2} \times \big(2 \times y^1\big)$

$=\,\,$ $\dfrac{1}{x^2+y^2} \times \big(2 \times y\big)$

$=\,\,$ $\dfrac{1}{x^2+y^2} \times \big(2y\big)$

$=\,\,$ $\dfrac{1}{x^2+y^2} \times 2y$

$=\,\,$ $\dfrac{1 \times 2y}{x^2+y^2}$

$=\,\,$ $\dfrac{2y}{x^2+y^2}$

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