What is the d/(dt)(1.1t+6.3t^3) ?

The d/(dt)(1.1t+6.3t^3) is 1.1+18.9t^2

What is the first d/(dt)(1.1t+6.3t^3) ?

The first d/(dt)(1.1t+6.3t^3) is 1.1+18.9t^2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{d}{d t} (1.1 t + 6.3 t^{3})

1.1 + 18.9 t^{2}

Solution steps

\frac{d}{d t} (1.1 t + 6.3 t^{3})

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (1.1 t) + \frac{d}{d t} (6.3 t^{3})

\frac{d}{d t} (1.1 t) = 1.1

\frac{d}{d t} (6.3 t^{3}) = 18.9 t^{2}

= 1.1 + 18.9 t^{2}

x \to \infty lim ((\frac{x + 1}{x + 2})^{\frac{- 4 x}{5}})

f (x) = \frac{x ^{3} + 7}{x}

\int \frac{3}{x ^{5}} d x

\int \frac{( 16 x ^{2} )}{( x ^{2} + 9 ) ^{5}} d x

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

What is the d/(dt)(1.1t+6.3t^3) ?
The d/(dt)(1.1t+6.3t^3) is 1.1+18.9t^2
What is the first d/(dt)(1.1t+6.3t^3) ?
The first d/(dt)(1.1t+6.3t^3) is 1.1+18.9t^2