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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

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

Solution steps

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

Treat

x

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

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