# Partial Derivative of log(x^2+y^2): Formula, Proof

The partial derivative of log(x^2+y^2) with respect to x is equal to 2x/(x2+y2) and with respect to y is equal to 2y/(x2+y2). So their formulas are as follows:

where ∂z/∂x is the partial derivative of z with respect to x.

## Partial Derivative of log(x2+y2) with respect to x

Answer: The partial derivative of log(x2+y2) w.r.t x is denoted by ∂/∂x[log(x2+y2)] and it is equal to 2x/(x2+y2).

Explanation:

To find the partial derivative of log(x2+y2) with respect to x, we will treat y as a constant and x as a variable. So we obtain that

$\dfrac{\partial }{\partial x}$ (log(x2+y2))

= $\dfrac{1}{x^2+y^2}\dfrac{\partial }{\partial x}$ (x2+y2), by the chain rule.

= $\dfrac{1}{x^2+y^2}$ (2x+0)

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

So the partial derivative of log(x2+y2) with respect to x is 2x/(x2+y2).

## Partial Derivative of log(x2+y2) with respect to y

Answer: The partial derivative of log(x2+y2) w.r.t y is denoted by ∂/∂y[log(x2+y2)] and it is equal to 2y/(x2+y2).

Explanation:

To find the partial derivative of log(x2+y2) with respect to y, we will treat x as a constant and y as a variable. Therefore,

$\dfrac{\partial }{\partial y}$ (log(x2+y2))

= $\dfrac{1}{x^2+y^2}\dfrac{\partial }{\partial y}$ (x2+y2)

= $\dfrac{1}{x^2+y^2}$ (0+2y)

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

So the partial derivative of log(x2+y2) with respect to y is 2y/(x2+y2).

Related Topics:

Partial Derivative of xy

## FAQs

### Q1: If z=log(x2+y2), then find ∂z/∂x.

Answer: If z=log(x2+y2), then by above ∂z/∂x = 2x/(x2+y2).

### Q2: If z=log(x2+y2), then find ∂z/∂y.

Answer: If z=log(x2+y2), then by above ∂z/∂y = 2y/(x2+y2).