What is the d/(dt)((x)(2x^2+1)) ?

The d/(dt)((x)(2x^2+1)) is 6t^2+1

What is the first d/(dt)((x)(2x^2+1)) ?

The first d/(dt)((x)(2x^2+1)) is 6t^2+1

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

6 t^{2} + 1

Solution steps

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = t, g = 2 t^{2} + 1

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

\frac{d t}{d t} = 1

\frac{d}{d t} (2 t^{2} + 1) = 4 t

= 1 \cdot (2 t^{2} + 1) + 4 tt

Simplify

1 \cdot (2 t^{2} + 1) + 4 tt : 6 t^{2} + 1

= 6 t^{2} + 1

\frac{2 ( 216 - 216 x + 72 x ^{2} - 8 x ^{3} )}{27}

\frac{x ^{3} + 16}{( x + 2 ) ^{2}}

\frac{A ( x - 2 ) + B}{( x - 2 ) ^{2}}

28

2 e^{- \frac{( x - 3 ) ^{2}}{2}} + 2 e^{- \frac{( x - 6 ) ^{2}}{2}}