What is the d/(dt)((x)(2e^{3t}+1)) ?

The d/(dt)((x)(2e^{3t}+1)) is 6e^{3t}x

What is the first d/(dt)((x)(2e^{3t}+1)) ?

The first d/(dt)((x)(2e^{3t}+1)) is 6e^{3t}x

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\frac{d}{d t} ((x) (2 e^{3 t} + 1))

6 e^{3 t} x

Solution steps

\frac{d}{d t} ((x) (2 e^{3 t} + 1))

Treat

x

as a constant

Take the constant out:

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

= x \frac{d}{d t} (2 e^{3 t} + 1)

Apply the Sum/Difference Rule:

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

= x (\frac{d}{d t} (2 e^{3 t}) + \frac{d}{d t} (1))

\frac{d}{d t} (2 e^{3 t}) = 6 e^{3 t}

\frac{d}{d t} (1) = 0

= x (6 e^{3 t} + 0)

Simplify

= 6 e^{3 t} x

\frac{( k - 2 ) ( k - 1 )}{( k + 2 ) ( k + 1 )}

- \frac{( x + 1 ) ( x - 2 ) ( x - 3 )}{5}

\frac{( a + b )}{( a - b ) ^{2}}

\frac{3 x - 4}{( 5 x + 2 ) ^{3}}

- 1 + 6 cos (\frac{2 π}{7} (x - 5))

What is the d/(dt)((x)(2e^{3t}+1)) ?
The d/(dt)((x)(2e^{3t}+1)) is 6e^{3t}x
What is the first d/(dt)((x)(2e^{3t}+1)) ?
The first d/(dt)((x)(2e^{3t}+1)) is 6e^{3t}x