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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{x ^{2} - y ^{2}}{x ( x ^{2} + y ^{2} )}

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d}{d x} (ln (sin (7 x)))

\frac{d}{d x} (- x^{2} - 2 x + 3)

\frac{d y}{d x} = (y + 1)^{- 1}

\frac{d}{d x} (4 e^{x} + \frac{8}{3 x})

\int_{1}^{8} \frac{x}{x ^{2} + 6 x + 13} d x

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