What is the derivative of ae^{bt} ?

The derivative of ae^{bt} is ae^{bt}b

What is the first derivative of ae^{bt} ?

The first derivative of ae^{bt} is ae^{bt}b

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derivative of

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

f (x) = \frac{3}{x + 2}, at x = 3

sin (9 x) + cos (2 x)

\int \frac{1 - z}{- 1 - z}

(ln (x))^{^{'}}

\int (x^{2} - 2 x - 3) d x

What is the derivative of ae^{bt} ?
The derivative of ae^{bt} is ae^{bt}b
What is the first derivative of ae^{bt} ?
The first derivative of ae^{bt} is ae^{bt}b