What is the d/(dt)(ae^{bt}) ?

The d/(dt)(ae^{bt}) is ae^{bt}b

What is the first d/(dt)(ae^{bt}) ?

The first d/(dt)(ae^{bt}) is ae^{bt}b

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\frac{d}{d t} (a e^{b t})

a e^{b t} b

Solution steps

\frac{d}{d t} (a e^{b t})

Take the constant out:

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

= a \frac{d}{d t} (e^{b t})

Apply the chain rule:

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

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

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

= a e^{b t} b

\frac{\partial}{\partial x} (e^{- t})

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

\frac{\partial}{\partial t} (sech^{2} (x - t))

\int 6 x^{4} e^{x^{5}} d x

\frac{d}{d x} ((2 x^{2} - 2 x) l e f (x, t) t (x^{2} - 3))