What is the derivative of ln|1+x-x^3| ?

The derivative of ln|1+x-x^3| is (-3x^2+1)/(1+x-x^3)

What is the first derivative of ln|1+x-x^3| ?

The first derivative of ln|1+x-x^3| is (-3x^2+1)/(1+x-x^3)

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\frac{d}{d x} (ln 1 + x - x^{3})

\frac{- 3 x ^{2} + 1}{1 + x - x ^{3}}

Solution steps

\frac{d}{d x} (ln 1 + x - x^{3})

Apply the chain rule:

\frac{1}{∣ 1 + x - x ^{3} ∣} \frac{d}{d x} (1 + x - x^{3})

= \frac{1}{∣ 1 + x - x ^{3} ∣} \frac{d}{d x} (1 + x - x^{3})

\frac{d}{d x} (1 + x - x^{3}) = \frac{( 1 + x - x ^{3} ) ( - 3 x ^{2} + 1 )}{∣ 1 + x - x ^{3} ∣}

= \frac{1}{∣ 1 + x - x ^{3} ∣} \cdot \frac{( 1 + x - x ^{3} ) ( - 3 x ^{2} + 1 )}{∣ 1 + x - x ^{3} ∣}

Simplify

\frac{1}{∣ 1 + x - x ^{3} ∣} \cdot \frac{( 1 + x - x ^{3} ) ( - 3 x ^{2} + 1 )}{∣ 1 + x - x ^{3} ∣} : \frac{- 3 x ^{2} + 1}{1 + x - x ^{3}}

= \frac{- 3 x ^{2} + 1}{1 + x - x ^{3}}

\int \frac{1}{x 1 + x ^{n}} d x

x \to \infty lim ((a)^{x})

y^{^{'}} + 3 x^{2} y = sin (x) e^{- x^{3}}, y (0) = 1

d er i v a t i v e x^{3} b^{5} + y^{4} b + z^{6}

\int_{0}^{10} 100 e^{0.1 x} d x

What is the derivative of ln|1+x-x^3| ?
The derivative of ln|1+x-x^3| is (-3x^2+1)/(1+x-x^3)
What is the first derivative of ln|1+x-x^3| ?
The first derivative of ln|1+x-x^3| is (-3x^2+1)/(1+x-x^3)