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

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

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

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

e^{3 t} \cdot 3

Solution steps

\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

= e^{3 t} \cdot 3

\frac{d}{d x} (a cos (t) + b sin (t))

y = 6 x, y = \frac{2}{3} x, y = 40 - x^{2}

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

\frac{\partial}{\partial x} (x^{2} \cdot cos (x y))

y = 10 x^{2}, y^{2} = \frac{1}{10} x