What is the d/(dt)((x)(4-8t^2)) ?

The d/(dt)((x)(4-8t^2)) is (dx)/(dt)(4-8t^2)-16tx

What is the first d/(dt)((x)(4-8t^2)) ?

The first d/(dt)((x)(4-8t^2)) is (dx)/(dt)(4-8t^2)-16tx

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\frac{d}{d t} ((x) (4 - 8 t^{2}))

\frac{d x}{d t} (4 - 8 t^{2}) - 16 t x

Solution steps

\frac{d}{d t} ((x) (4 - 8 t^{2}))

Apply the Product Rule:

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

f = x, g = 4 - 8 t^{2}

= \frac{d x}{d t} (4 - 8 t^{2}) + \frac{d}{d t} (4 - 8 t^{2}) x

\frac{d}{d t} (4 - 8 t^{2}) = - 16 t

= \frac{d x}{d t} (4 - 8 t^{2}) + (- 16 t) x

Simplify

= \frac{d x}{d t} (4 - 8 t^{2}) - 16 t x

A (2, - 3) y B (8, 1)

\frac{( x ^{2} + 4 ) 9 - x ^{2}}{x ^{3} - 1}

x (x d v + v d x)

n o t a v e i s (- 5 + \frac{x ^{2}}{2})^{3}

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

What is the d/(dt)((x)(4-8t^2)) ?
The d/(dt)((x)(4-8t^2)) is (dx)/(dt)(4-8t^2)-16tx
What is the first d/(dt)((x)(4-8t^2)) ?
The first d/(dt)((x)(4-8t^2)) is (dx)/(dt)(4-8t^2)-16tx