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

Function | Partial Derivative |

z=log(x^{2}+y^{2}) | ∂z/∂x = 2x/(x^{2}+y^{2}) |

z=log(x^{2}+y^{2}) | ∂z/∂y = 2y/(x^{2}+y^{2}) |

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

## Partial Derivative of log(x^{2}+y^{2}) with respect to x

**Answer:** The partial derivative of log(x^{2}+y^{2}) w.r.t x is denoted by ∂/∂x[log(x^{2}+y^{2})] and it is equal to 2x/(x^{2}+y^{2}).

**Explanation:**

To find the partial derivative of log(x^{2}+y^{2}) 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(x^{2}+y^{2}))

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

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

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

So the partial derivative of log(x^{2}+y^{2}) with respect to x is 2x/(x^{2}+y^{2}).

Remark: If z=log(x^{2}+y^{2}), then by above ∂z/∂x = 2x/(x^{2}+y^{2}). |

## Partial Derivative of log(x^{2}+y^{2}) with respect to y

**Answer:** The partial derivative of log(x^{2}+y^{2}) w.r.t y is denoted by ∂/∂y[log(x^{2}+y^{2})] and it is equal to 2y/(x^{2}+y^{2}).

**Explanation:**

To find the partial derivative of log(x^{2}+y^{2}) with respect to y, we will treat x as a constant and y as a variable. Therefore,

$\dfrac{\partial }{\partial y}$ (log(x^{2}+y^{2}))

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

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

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

So the partial derivative of log(x^{2}+y^{2}) with respect to y is 2y/(x^{2}+y^{2}).

**Related Topics:**

## FAQs

### Q1: If z=log(x^{2}+y^{2}), then find ∂z/∂x.

Answer: If z=log(x^{2}+y^{2}), then by above ∂z/∂x = 2x/(x^{2}+y^{2}).

### Q2: If z=log(x^{2}+y^{2}), then find ∂z/∂y.

Answer: If z=log(x^{2}+y^{2}), then by above ∂z/∂y = 2y/(x^{2}+y^{2}).