What is the d/(dt)((x)(t^2+5)) ?

The d/(dt)((x)(t^2+5)) is (dx)/(dt)(t^2+5)+2tx

What is the first d/(dt)((x)(t^2+5)) ?

The first d/(dt)((x)(t^2+5)) is (dx)/(dt)(t^2+5)+2tx

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\frac{d}{d t} ((x) (t^{2} + 5))

\frac{d x}{d t} (t^{2} + 5) + 2 t x

Solution steps

\frac{d}{d t} ((x) (t^{2} + 5))

Apply the Product Rule:

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

f = x, g = t^{2} + 5

= \frac{d x}{d t} (t^{2} + 5) + \frac{d}{d t} (t^{2} + 5) x

\frac{d}{d t} (t^{2} + 5) = 2 t

= \frac{d x}{d t} (t^{2} + 5) + 2 t x

(5 sin (θ) + 2)^{2}

(tan (β) + 1)^{2}

(1 - a) \cdot (1 - b) l e f t (1 - c)

(e^{- a s} - 1)^{2}

(2 + 2 i x - 2 x)^{2}

What is the d/(dt)((x)(t^2+5)) ?
The d/(dt)((x)(t^2+5)) is (dx)/(dt)(t^2+5)+2tx
What is the first d/(dt)((x)(t^2+5)) ?
The first d/(dt)((x)(t^2+5)) is (dx)/(dt)(t^2+5)+2tx