What is the d/(dt)((x)(t^3+t-3)) ?

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

What is the first d/(dt)((x)(t^3+t-3)) ?

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

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\frac{d}{d t} ((x) (t^{3} + t - 3))

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

Solution steps

\frac{d}{d t} ((x) (t^{3} + t - 3))

Apply the Product Rule:

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

f = x, g = t^{3} + t - 3

= \frac{d x}{d t} (t^{3} + t - 3) + \frac{d}{d t} (t^{3} + t - 3) x

\frac{d}{d t} (t^{3} + t - 3) = 3 t^{2} + 1

= \frac{d x}{d t} (t^{3} + t - 3) + (3 t^{2} + 1) x

\frac{1}{5} (1 + e^{2 x})

(x + 3) \cdot ln (x + 3)

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

\frac{( 1 + r ) ( 1 - a ) ( x ) + c}{( 1 + r )}

2 x^{2} - x^{- 1}

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