What is (\partial)/(\partial y)(1/2 ln(x^2+y^2)) ?

The answer to (\partial)/(\partial y)(1/2 ln(x^2+y^2)) is y/(x^2+y^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

(\partial)/(\partial y)(1/2 ln(x^2+y^2))

\frac{\partial}{\partial y} (\frac{1}{2} ln (x^{2} + y^{2}))

\frac{y}{x ^{2} + y ^{2}}

Solution steps

\frac{\partial}{\partial y} (\frac{1}{2} ln (x^{2} + y^{2}))

Treat

x

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{2} \frac{\partial}{\partial y} (ln (x^{2} + y^{2}))

Apply the chain rule:

\frac{1}{x ^{2} + y ^{2}} \frac{\partial}{\partial y} (x^{2} + y^{2})

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

\frac{\partial}{\partial y} (x^{2} + y^{2}) = 2 y

= \frac{1}{2} \cdot \frac{1}{x ^{2} + y ^{2}} \cdot 2 y

Simplify

\frac{1}{2} \cdot \frac{1}{x ^{2} + y ^{2}} \cdot 2 y : \frac{y}{x ^{2} + y ^{2}}

= \frac{y}{x ^{2} + y ^{2}}

What is (\partial)/(\partial y)(1/2 ln(x^2+y^2)) ?
The answer to (\partial)/(\partial y)(1/2 ln(x^2+y^2)) is y/(x^2+y^2)