What is the derivative of f(x)=ln(1-x^2) ?

The derivative of f(x)=ln(1-x^2) is -(2x)/(1-x^2)

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

The first derivative of f(x)=ln(1-x^2) is -(2x)/(1-x^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

f (x) = ln (1 - x^{2})

- \frac{2 x}{1 - x ^{2}}

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

= - \frac{2 x}{1 - x ^{2}}

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

n = 1 \sum \infty 2 0^{n} x^{n} n!

\int_{- π}^{2 π} sin^{2} (x) d x

e^{6}

\int \frac{16}{16 - x ^{2}} d x

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