What is the d/(dt)(b(t)ln(t)+ct^2) ?

The d/(dt)(b(t)ln(t)+ct^2) is b(ln(t)+1)+2ct

What is the first d/(dt)(b(t)ln(t)+ct^2) ?

The first d/(dt)(b(t)ln(t)+ct^2) is b(ln(t)+1)+2ct

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{d}{d t} (b (t) ln (t) + c t^{2})

b (ln (t) + 1) + 2 c t

Solution steps

\frac{d}{d t} (b (t) ln (t) + c t^{2})

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (b t ln (t)) + \frac{d}{d t} (c t^{2})

\frac{d}{d t} (b t ln (t)) = b (ln (t) + 1)

\frac{d}{d t} (c t^{2}) = 2 c t

= b (ln (t) + 1) + 2 c t

\int \frac{x ^{2} + 6}{x x ^{2} - 9} d x

\frac{d}{d x} (\frac{e ^{x} \cdot cos ( x )}{e ^{x} - 1})

\frac{\partial}{\partial y} ((2 x - 3 y)^{3})

x \to + (- 8) lim (2)

x^{2} + x - 2, 0

What is the d/(dt)(b(t)ln(t)+ct^2) ?
The d/(dt)(b(t)ln(t)+ct^2) is b(ln(t)+1)+2ct
What is the first d/(dt)(b(t)ln(t)+ct^2) ?
The first d/(dt)(b(t)ln(t)+ct^2) is b(ln(t)+1)+2ct