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

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

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d/(dt)(2e^{3t})

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

6 e^{3 t}

Solution steps

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

Take the constant out:

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

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

Apply the chain rule:

e^{3 t} \frac{d}{d t} (3 t)

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

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

= 2 e^{3 t} \cdot 3

Simplify

= 6 e^{3 t}

\frac{d}{d x} (\frac{sin ( x )}{7 + cos ( x )})

\frac{d}{d x} (5 + 6 x^{\frac{3}{2}})

x^{^{''}} + 4 x = 3

\int \frac{7}{x ^{3} - 27} d x

x \to 4 lim (- \frac{3}{( x - 4 ) ^{2}})