Evaluate Partial Derivative of $\log_{e}{(x^2+y^2)}$ with respect to $x$ and $y$ by Derivative Rules

Math Doubts

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.

Partial Derivative with respect to x

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.

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

Find the Partial Derivative by Chain Rule

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 $x$.

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

Distribute the Partial derivative over addition

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.

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

Simplify the mathematical expression

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

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

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

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

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

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

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

Partial Derivative with respect to y

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

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

Find the Partial Derivative by Chain Rule

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 $y$.

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

Distribute the Partial derivative over addition

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 $y$.

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