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) \cdot 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)

(3 - sin (x))^{2}

3 x^{\frac{1}{3}} (x + 2)

x (3 + 5 x)

\frac{d}{d y} ((x) (x^{y} + \frac{1}{x ^{2}}))

(4 e - 2 x)^{3}

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)