What is the d/(dt)((x)((t^3e^{2(-3+4t)}))) ?

The d/(dt)((x)((t^3e^{2(-3+4t)}))) is x(3t^2e^{2(-3+4t)}+e^{2(-3+4t)}*8t^3)

What is the first d/(dt)((x)((t^3e^{2(-3+4t)}))) ?

The first d/(dt)((x)((t^3e^{2(-3+4t)}))) is x(3t^2e^{2(-3+4t)}+e^{2(-3+4t)}*8t^3)

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\frac{d}{d t} ((x) ((t^{3} e^{2 (- 3 + 4 t)})))

x (3 t^{2} e^{2 (- 3 + 4 t)} + e^{2 (- 3 + 4 t)} \cdot 8 t^{3})

Solution steps

\frac{d}{d t} ((x) ((t^{3} e^{2 (- 3 + 4 t)})))

Treat

x

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= x \frac{d}{d t} (t^{3} e^{2 (- 3 + 4 t)})

Apply the Product Rule:

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

f = t^{3}, g = e^{2 (- 3 + 4 t)}

= x (\frac{d}{d t} (t^{3}) e^{2 (- 3 + 4 t)} + \frac{d}{d t} (e^{2 (- 3 + 4 t)}) t^{3})

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

\frac{d}{d t} (e^{2 (- 3 + 4 t)}) = e^{2 (- 3 + 4 t)} \cdot 8

= x (3 t^{2} e^{2 (- 3 + 4 t)} + e^{2 (- 3 + 4 t)} \cdot 8 t^{3})

\frac{( - 4 y + 1 ) ( 12 x ^{2} - 8 x y - 6 x + 16 y ^{2} + 4 y + 1 )}{8}

(9 x - 2) e^{5 x}

\frac{( - 10 x ^{2} + 160 x )}{( 5 x ^{2} - 2 x + 16 ) ^{2}}

(a + b \cdot cos (t))^{3}

What is the d/(dt)((x)((t^3e^{2(-3+4t)}))) ?
The d/(dt)((x)((t^3e^{2(-3+4t)}))) is x(3t^2e^{2(-3+4t)}+e^{2(-3+4t)}*8t^3)
What is the first d/(dt)((x)((t^3e^{2(-3+4t)}))) ?
The first d/(dt)((x)((t^3e^{2(-3+4t)}))) is x(3t^2e^{2(-3+4t)}+e^{2(-3+4t)}*8t^3)