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

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

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

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

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derivative of

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}}

\int (sin (2 x) + cos (2 x))^{2} d x

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

y = 3 - x, y = 2 x^{2}, [0, 1]

f (x) = 2 x^{2} + 3 x, (- 2, 2)

\frac{\partial}{\partial x} (x^{3} - 12 x + y^{3} - 27 y)

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