What is the d/(dt)((B)(B+Bt^2)) ?

The d/(dt)((B)(B+Bt^2)) is 2B^2t

What is the first d/(dt)((B)(B+Bt^2)) ?

The first d/(dt)((B)(B+Bt^2)) is 2B^2t

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\frac{d}{d t} ((B) (B + B t^{2}))

2 B^{2} t

Solution steps

\frac{d}{d t} ((B) (B + B t^{2}))

Treat

B

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= B \frac{d}{d t} (B + B t^{2})

Apply the Sum/Difference Rule:

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

= B (\frac{d B}{d t} + \frac{d}{d t} (B t^{2}))

\frac{d B}{d t} = 0

\frac{d}{d t} (B t^{2}) = 2 Bt

= B (0 + 2 Bt)

Simplify

B (0 + 2 Bt) : 2 B^{2} t

= 2 B^{2} t

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