What is the derivative of 6+(2t)/(1+t^3)-e^{-0.02t} ?

The derivative of 6+(2t)/(1+t^3)-e^{-0.02t} is 0.02e^{-0.02t}+(2(-2t^3+1))/((t^3+1)^2)

What is the first derivative of 6+(2t)/(1+t^3)-e^{-0.02t} ?

The first derivative of 6+(2t)/(1+t^3)-e^{-0.02t} is 0.02e^{-0.02t}+(2(-2t^3+1))/((t^3+1)^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

6 + \frac{2 t}{1 + t ^{3}} - e^{- 0.02 t}

0.02 e^{- 0.02 t} + \frac{2 ( - 2 t ^{3} + 1 )}{( t ^{3} + 1 ) ^{2}}

Solution steps

\frac{d}{d t} (6 + \frac{2 t}{1 + t ^{3}} - e^{- 0.02 t})

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d t} (6) + \frac{d}{d t} (\frac{2 t}{1 + t ^{3}}) - \frac{d}{d t} (e^{- 0.02 t})

\frac{d}{d t} (6) = 0

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

\frac{d}{d t} (e^{- 0.02 t}) = - 0.02 e^{- 0.02 t}

= 0 + \frac{2 ( 1 - 2 t ^{3} )}{( 1 + t ^{3} ) ^{2}} - (- 0.02 e^{- 0.02 t})

Simplify

= 0.02 e^{- 0.02 t} + \frac{2 ( - 2 t ^{3} + 1 )}{( t ^{3} + 1 ) ^{2}}

\frac{d}{d s} (e^{- s})

\int 5 cos (\frac{x}{2}) d x

x^{2} 16 - x^{2}

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

- \frac{8}{x ^{2}}

What is the derivative of 6+(2t)/(1+t^3)-e^{-0.02t} ?
The derivative of 6+(2t)/(1+t^3)-e^{-0.02t} is 0.02e^{-0.02t}+(2(-2t^3+1))/((t^3+1)^2)
What is the first derivative of 6+(2t)/(1+t^3)-e^{-0.02t} ?
The first derivative of 6+(2t)/(1+t^3)-e^{-0.02t} is 0.02e^{-0.02t}+(2(-2t^3+1))/((t^3+1)^2)