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

The d/(dt)((θ)(2e^{2t}-4e^{3t})) is θ(4e^{2t}-12e^{3t})

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

The first d/(dt)((θ)(2e^{2t}-4e^{3t})) is θ(4e^{2t}-12e^{3t})

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

θ (4 e^{2 t} - 12 e^{3 t})

Solution steps

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

Treat

θ

as a constant

Take the constant out:

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

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

\frac{d}{d t} (2 e^{2 t}) = 4 e^{2 t}

\frac{d}{d t} (4 e^{3 t}) = 12 e^{3 t}

= θ (4 e^{2 t} - 12 e^{3 t})

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

- (x^{3} + 3 y^{2}) d y

3 (3 - x) e^{x}

l im x \Rightarrow 0 + (1 + x) 7 x

\frac{s ^{3} - 7 s ^{2} + 19 s - 17}{( s - 2 ) ^{2} ( s - 3 ) ^{3}}

What is the d/(dt)((θ)(2e^{2t}-4e^{3t})) ?
The d/(dt)((θ)(2e^{2t}-4e^{3t})) is θ(4e^{2t}-12e^{3t})
What is the first d/(dt)((θ)(2e^{2t}-4e^{3t})) ?
The first d/(dt)((θ)(2e^{2t}-4e^{3t})) is θ(4e^{2t}-12e^{3t})