What is the derivative of y=ln|5+t-t^3| ?

The derivative of y=ln|5+t-t^3| is (-3t^2+1)/(5+t-t^3)

What is the first derivative of y=ln|5+t-t^3| ?

The first derivative of y=ln|5+t-t^3| is (-3t^2+1)/(5+t-t^3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

y = ln 5 + t - t^{3}

\frac{- 3 t ^{2} + 1}{5 + t - t ^{3}}

Solution steps

\frac{d}{d t} (ln 5 + t - t^{3})

Apply the chain rule:

\frac{1}{∣ 5 + t - t ^{3} ∣} \frac{d}{d t} (5 + t - t^{3})

= \frac{1}{∣ 5 + t - t ^{3} ∣} \frac{d}{d t} (5 + t - t^{3})

\frac{d}{d t} (5 + t - t^{3}) = \frac{( 5 + t - t ^{3} ) ( - 3 t ^{2} + 1 )}{∣ 5 + t - t ^{3} ∣}

= \frac{1}{∣ 5 + t - t ^{3} ∣} \cdot \frac{( 5 + t - t ^{3} ) ( - 3 t ^{2} + 1 )}{∣ 5 + t - t ^{3} ∣}

Simplify

\frac{1}{∣ 5 + t - t ^{3} ∣} \cdot \frac{( 5 + t - t ^{3} ) ( - 3 t ^{2} + 1 )}{∣ 5 + t - t ^{3} ∣} : \frac{- 3 t ^{2} + 1}{5 + t - t ^{3}}

= \frac{- 3 t ^{2} + 1}{5 + t - t ^{3}}

\int (5 cos (x) - 4 sin (x)) d x

\frac{1}{x - 2}

f (x) = e^{4 x}

\frac{d ^{2}}{d x ^{2}} (x^{\frac{8}{9}} + 5 x)

\int_{3}^{\infty} \frac{1}{x} d x

What is the derivative of y=ln|5+t-t^3| ?
The derivative of y=ln|5+t-t^3| is (-3t^2+1)/(5+t-t^3)
What is the first derivative of y=ln|5+t-t^3| ?
The first derivative of y=ln|5+t-t^3| is (-3t^2+1)/(5+t-t^3)