What is the d/(dt)((x)(1+4t-t^2)) ?

The d/(dt)((x)(1+4t-t^2)) is (dx)/(dt)(1+4t-t^2)+(-2t+4)x

What is the first d/(dt)((x)(1+4t-t^2)) ?

The first d/(dt)((x)(1+4t-t^2)) is (dx)/(dt)(1+4t-t^2)+(-2t+4)x

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\frac{d}{d t} ((x) (1 + 4 t - t^{2}))

\frac{d x}{d t} (1 + 4 t - t^{2}) + (- 2 t + 4) x

Solution steps

\frac{d}{d t} ((x) (1 + 4 t - t^{2}))

Apply the Product Rule:

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

f = x, g = 1 + 4 t - t^{2}

= \frac{d x}{d t} (1 + 4 t - t^{2}) + \frac{d}{d t} (1 + 4 t - t^{2}) x

\frac{d}{d t} (1 + 4 t - t^{2}) = - 2 t + 4

= \frac{d x}{d t} (1 + 4 t - t^{2}) + (- 2 t + 4) x

(e^{- y} + 2)^{2}

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

(1 - 7) + (7 - 7)

\frac{( s + 3 )}{( s + 2.5 ) ( s + 4 ) ^{2}}

\frac{( s + 1 ) ( s - 2 )}{( s + 3 )}

What is the d/(dt)((x)(1+4t-t^2)) ?
The d/(dt)((x)(1+4t-t^2)) is (dx)/(dt)(1+4t-t^2)+(-2t+4)x
What is the first d/(dt)((x)(1+4t-t^2)) ?
The first d/(dt)((x)(1+4t-t^2)) is (dx)/(dt)(1+4t-t^2)+(-2t+4)x