What is the derivative of y=ln((x^2-1)/(x^2+1)) ?

The derivative of y=ln((x^2-1)/(x^2+1)) is (4x)/((x^2-1)(x^2+1))

What is the first derivative of y=ln((x^2-1)/(x^2+1)) ?

The first derivative of y=ln((x^2-1)/(x^2+1)) is (4x)/((x^2-1)(x^2+1))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

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

\frac{4 x}{( x ^{2} - 1 ) ( x ^{2} + 1 )}

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d x} (\frac{x ^{2} - 1}{x ^{2} + 1}) = \frac{4 x}{( x ^{2} + 1 ) ^{2}}

= \frac{1}{\frac{x ^{2} - 1}{x ^{2} + 1}} \cdot \frac{4 x}{( x ^{2} + 1 ) ^{2}}

Simplify

\frac{1}{\frac{x ^{2} - 1}{x ^{2} + 1}} \cdot \frac{4 x}{( x ^{2} + 1 ) ^{2}} : \frac{4 x}{( x ^{2} - 1 ) ( x ^{2} + 1 )}

= \frac{4 x}{( x ^{2} - 1 ) ( x ^{2} + 1 )}

\frac{d y}{d x} + 12 y = 6

- 10 e^{- 5 x^{2}} x

\int 6 x ln (5 x) d x

y = x^{4} - 4 x^{2} + 4, y = x^{2}

f (x) = 2 x - x^{2}, (a, 0)

What is the derivative of y=ln((x^2-1)/(x^2+1)) ?
The derivative of y=ln((x^2-1)/(x^2+1)) is (4x)/((x^2-1)(x^2+1))
What is the first derivative of y=ln((x^2-1)/(x^2+1)) ?
The first derivative of y=ln((x^2-1)/(x^2+1)) is (4x)/((x^2-1)(x^2+1))