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Derivative of $$$e^{4 t}$$$
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Find $$$\frac{d}{dt} \left(e^{4 t}\right)$$$.
Solution
The function $$$e^{4 t}$$$ is the composition $$$f{\left(g{\left(t \right)} \right)}$$$ of two functions $$$f{\left(u \right)} = e^{u}$$$ and $$$g{\left(t \right)} = 4 t$$$.
Apply the chain rule $$$\frac{d}{dt} \left(f{\left(g{\left(t \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dt} \left(g{\left(t \right)}\right)$$$:
$${\color{red}\left(\frac{d}{dt} \left(e^{4 t}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(e^{u}\right) \frac{d}{dt} \left(4 t\right)\right)}$$The derivative of the exponential is $$$\frac{d}{du} \left(e^{u}\right) = e^{u}$$$:
$${\color{red}\left(\frac{d}{du} \left(e^{u}\right)\right)} \frac{d}{dt} \left(4 t\right) = {\color{red}\left(e^{u}\right)} \frac{d}{dt} \left(4 t\right)$$Return to the old variable:
$$e^{{\color{red}\left(u\right)}} \frac{d}{dt} \left(4 t\right) = e^{{\color{red}\left(4 t\right)}} \frac{d}{dt} \left(4 t\right)$$Apply the constant multiple rule $$$\frac{d}{dt} \left(c f{\left(t \right)}\right) = c \frac{d}{dt} \left(f{\left(t \right)}\right)$$$ with $$$c = 4$$$ and $$$f{\left(t \right)} = t$$$:
$$e^{4 t} {\color{red}\left(\frac{d}{dt} \left(4 t\right)\right)} = e^{4 t} {\color{red}\left(4 \frac{d}{dt} \left(t\right)\right)}$$Apply the power rule $$$\frac{d}{dt} \left(t^{n}\right) = n t^{n - 1}$$$ with $$$n = 1$$$, in other words, $$$\frac{d}{dt} \left(t\right) = 1$$$:
$$4 e^{4 t} {\color{red}\left(\frac{d}{dt} \left(t\right)\right)} = 4 e^{4 t} {\color{red}\left(1\right)}$$Thus, $$$\frac{d}{dt} \left(e^{4 t}\right) = 4 e^{4 t}$$$.
Answer
$$$\frac{d}{dt} \left(e^{4 t}\right) = 4 e^{4 t}$$$A