What is the d/(dt)((x)(t^2+6)) ?

The d/(dt)((x)(t^2+6)) is (dx)/(dt)(t^2+6)+2tx

What is the first d/(dt)((x)(t^2+6)) ?

The first d/(dt)((x)(t^2+6)) is (dx)/(dt)(t^2+6)+2tx

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\frac{d}{d t} ((x) (t^{2} + 6))

\frac{d x}{d t} (t^{2} + 6) + 2 t x

Solution steps

\frac{d}{d t} ((x) (t^{2} + 6))

Apply the Product Rule:

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

f = x, g = t^{2} + 6

= \frac{d x}{d t} (t^{2} + 6) + \frac{d}{d t} (t^{2} + 6) x

\frac{d}{d t} (t^{2} + 6) = 2 t

= \frac{d x}{d t} (t^{2} + 6) + 2 t x

\frac{\frac{d}{d x} ( 1 + x ( lo g _{10} ( x ) + 1 ) )}{x}

\frac{x ( x + 1 ) ( x + 55 )}{( x - 5 ) ( x + 5 ) ( 5 x + 3 )}

145447 \cdot (1 - ((- 98.07089)^{0.234957}))

\frac{x ^{2} + 15}{( x + 3 ) ( x ^{2} + 3 )}

π (- 4 cos (x) + 4)

What is the d/(dt)((x)(t^2+6)) ?
The d/(dt)((x)(t^2+6)) is (dx)/(dt)(t^2+6)+2tx
What is the first d/(dt)((x)(t^2+6)) ?
The first d/(dt)((x)(t^2+6)) is (dx)/(dt)(t^2+6)+2tx