What is the first d/(du)((x)(x+1)) ?

The first d/(du)((x)(x+1)) is 2u+1

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d/(du)((x)(x+1))

\frac{d}{d u} ((u) (u + 1))

2 u + 1

Solution steps

\frac{d}{d u} ((u) (u + 1))

Apply the Product Rule:

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

f = u, g = u + 1

= \frac{d u}{d u} (u + 1) + \frac{d}{d u} (u + 1) u

\frac{d u}{d u} = 1

\frac{d}{d u} (u + 1) = 1

= 1 \cdot (u + 1) + 1 \cdot u

Simplify

1 \cdot (u + 1) + 1 \cdot u : 2 u + 1

= 2 u + 1

2 (2 x - 4)

440 2 + 2 (2 - 2)

(f x + e) \cdot (x^{2} - x + 1) \cdot (x - 1)

- sin (x) (sin (x) + 1) - cos^{2} (x)

2 t (t^{2} - sin (x) + 1)