Derivative of $$$9 t^{2} + 4$$$
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Your Input
Find $$$\frac{d}{dt} \left(9 t^{2} + 4\right)$$$.
Solution
The derivative of a sum/difference is the sum/difference of derivatives:
$${\color{red}\left(\frac{d}{dt} \left(9 t^{2} + 4\right)\right)} = {\color{red}\left(\frac{d}{dt} \left(9 t^{2}\right) + \frac{d}{dt} \left(4\right)\right)}$$The derivative of a constant is $$$0$$$:
$${\color{red}\left(\frac{d}{dt} \left(4\right)\right)} + \frac{d}{dt} \left(9 t^{2}\right) = {\color{red}\left(0\right)} + \frac{d}{dt} \left(9 t^{2}\right)$$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 $$$c = 9$$$ and $$$f{\left(t \right)} = t^{2}$$$:
$${\color{red}\left(\frac{d}{dt} \left(9 t^{2}\right)\right)} = {\color{red}\left(9 \frac{d}{dt} \left(t^{2}\right)\right)}$$Apply the power rule $$$\frac{d}{dt} \left(t^{n}\right) = n t^{n - 1}$$$ with $$$n = 2$$$:
$$9 {\color{red}\left(\frac{d}{dt} \left(t^{2}\right)\right)} = 9 {\color{red}\left(2 t\right)}$$Thus, $$$\frac{d}{dt} \left(9 t^{2} + 4\right) = 18 t$$$.
Answer
$$$\frac{d}{dt} \left(9 t^{2} + 4\right) = 18 t$$$A