What is the d/(dt)((x)(4(x^2+1))) ?

The d/(dt)((x)(4(x^2+1))) is 4(3t^2+1)

What is the first d/(dt)((x)(4(x^2+1))) ?

The first d/(dt)((x)(4(x^2+1))) is 4(3t^2+1)

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

4 (3 t^{2} + 1)

Solution steps

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

Take the constant out:

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

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

Apply the Product Rule:

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

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

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

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

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

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

Simplify

4 (1 \cdot (t^{2} + 1) + 2 tt) : 4 (3 t^{2} + 1)

= 4 (3 t^{2} + 1)

- \frac{1}{- 2 e ^{- x} ( x + 1 ) + C}

x^{2} + 1^{2}

- 2 - 7 k (k = - 1)

(x - 1) e^{3}

(2 d t + 2 d w) \cdot (3 d t - d w)

What is the d/(dt)((x)(4(x^2+1))) ?
The d/(dt)((x)(4(x^2+1))) is 4(3t^2+1)
What is the first d/(dt)((x)(4(x^2+1))) ?
The first d/(dt)((x)(4(x^2+1))) is 4(3t^2+1)