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Derivative of $$$\sin^{3}{\left(t \right)}$$$
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Your Input
Find $$$\frac{d}{dt} \left(\sin^{3}{\left(t \right)}\right)$$$.
Solution
The function $$$\sin^{3}{\left(t \right)}$$$ is the composition $$$f{\left(g{\left(t \right)} \right)}$$$ of two functions $$$f{\left(u \right)} = u^{3}$$$ and $$$g{\left(t \right)} = \sin{\left(t \right)}$$$.
Apply the chain rule $$$\frac{d}{dt} \left(f{\left(g{\left(t \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dt} \left(g{\left(t \right)}\right)$$$:
$${\color{red}\left(\frac{d}{dt} \left(\sin^{3}{\left(t \right)}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(u^{3}\right) \frac{d}{dt} \left(\sin{\left(t \right)}\right)\right)}$$Apply the power rule $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$ with $$$n = 3$$$:
$${\color{red}\left(\frac{d}{du} \left(u^{3}\right)\right)} \frac{d}{dt} \left(\sin{\left(t \right)}\right) = {\color{red}\left(3 u^{2}\right)} \frac{d}{dt} \left(\sin{\left(t \right)}\right)$$Return to the old variable:
$$3 {\color{red}\left(u\right)}^{2} \frac{d}{dt} \left(\sin{\left(t \right)}\right) = 3 {\color{red}\left(\sin{\left(t \right)}\right)}^{2} \frac{d}{dt} \left(\sin{\left(t \right)}\right)$$The derivative of the sine is $$$\frac{d}{dt} \left(\sin{\left(t \right)}\right) = \cos{\left(t \right)}$$$:
$$3 \sin^{2}{\left(t \right)} {\color{red}\left(\frac{d}{dt} \left(\sin{\left(t \right)}\right)\right)} = 3 \sin^{2}{\left(t \right)} {\color{red}\left(\cos{\left(t \right)}\right)}$$Thus, $$$\frac{d}{dt} \left(\sin^{3}{\left(t \right)}\right) = 3 \sin^{2}{\left(t \right)} \cos{\left(t \right)}$$$.
Answer
$$$\frac{d}{dt} \left(\sin^{3}{\left(t \right)}\right) = 3 \sin^{2}{\left(t \right)} \cos{\left(t \right)}$$$A