What is the derivative of ln(((x+y)/((x-y)))) ?

The derivative of ln(((x+y)/((x-y)))) is (2x\frac{dy)/(dx)-2y}{(x+y)(x-y)}

What is the first derivative of ln(((x+y)/((x-y)))) ?

The first derivative of ln(((x+y)/((x-y)))) is (2x\frac{dy)/(dx)-2y}{(x+y)(x-y)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{d}{d x} (ln (\frac{( x + y )}{( x - y )}))

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

Solution steps

\frac{d}{d x} (ln (\frac{( x + y )}{( x - y )}))

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d}{d x} (e^{x^{9}})

\frac{d}{d x} (- 5 e^{- 2.5 x^{2}} x)

y^{^{'}} = \frac{x - 1}{y - 3}

\int \frac{x ln ( 1 + x ^{2} )}{1 + x ^{2}} d x

\int_{0}^{l n (2)} x e^{x} d x

What is the derivative of ln(((x+y)/((x-y)))) ?
The derivative of ln(((x+y)/((x-y)))) is (2x\frac{dy)/(dx)-2y}{(x+y)(x-y)}
What is the first derivative of ln(((x+y)/((x-y)))) ?
The first derivative of ln(((x+y)/((x-y)))) is (2x\frac{dy)/(dx)-2y}{(x+y)(x-y)}