What is the d/(dt)((3e^{3αt}xα)/4) ?

The d/(dt)((3e^{3αt}xα)/4) is (9e^{3αt}α^2x)/4

What is the first d/(dt)((3e^{3αt}xα)/4) ?

The first d/(dt)((3e^{3αt}xα)/4) is (9e^{3αt}α^2x)/4

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\frac{d}{d t} (\frac{3 e ^{3 α t} xα}{4})

\frac{9 e ^{3 α t} α ^{2} x}{4}

Solution steps

\frac{d}{d t} (\frac{3 e ^{3 α t} xα}{4})

Treat

α, x

as constants

Take the constant out:

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

= \frac{3 xα}{4} \frac{d}{d t} (e^{3 α t})

Apply the chain rule:

e^{3 α t} \frac{d}{d t} (3 α t)

= e^{3 α t} \frac{d}{d t} (3 α t)

\frac{d}{d t} (3 α t) = 3 α

= \frac{3 xα}{4} e^{3 α t} \cdot 3 α

Simplify

\frac{3 xα}{4} e^{3 α t} \cdot 3 α : \frac{9 e ^{3 α t} α ^{2} x}{4}

= \frac{9 e ^{3 α t} α ^{2} x}{4}

What is the d/(dt)((3e^{3αt}xα)/4) ?
The d/(dt)((3e^{3αt}xα)/4) is (9e^{3αt}α^2x)/4
What is the first d/(dt)((3e^{3αt}xα)/4) ?
The first d/(dt)((3e^{3αt}xα)/4) is (9e^{3αt}α^2x)/4