What is the d/(dt)(ln|1+t-t^3|) ?

The d/(dt)(ln|1+t-t^3|) is (-3t^2+1)/(1+t-t^3)

What is the first d/(dt)(ln|1+t-t^3|) ?

The first d/(dt)(ln|1+t-t^3|) is (-3t^2+1)/(1+t-t^3)

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

x \to 6 - lim (∣ x - 6 ∣)

\int (3 e^{- 3 x}) d x

\frac{\partial}{\partial y} (- x sin (x y))

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

\frac{\partial}{\partial x} (cos (\frac{x}{y}))

What is the d/(dt)(ln|1+t-t^3|) ?
The d/(dt)(ln|1+t-t^3|) is (-3t^2+1)/(1+t-t^3)
What is the first d/(dt)(ln|1+t-t^3|) ?
The first d/(dt)(ln|1+t-t^3|) is (-3t^2+1)/(1+t-t^3)