Derivative of $$$\cos{\left(3 x \right)}$$$
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Solution
The function $$$\cos{\left(3 x \right)}$$$ is the composition $$$f{\left(g{\left(x \right)} \right)}$$$ of two functions $$$f{\left(u \right)} = \cos{\left(u \right)}$$$ and $$$g{\left(x \right)} = 3 x$$$.
Apply the chain rule $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:
$${\color{red}\left(\frac{d}{dx} \left(\cos{\left(3 x \right)}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\cos{\left(u \right)}\right) \frac{d}{dx} \left(3 x\right)\right)}$$The derivative of the cosine is $$$\frac{d}{du} \left(\cos{\left(u \right)}\right) = - \sin{\left(u \right)}$$$:
$${\color{red}\left(\frac{d}{du} \left(\cos{\left(u \right)}\right)\right)} \frac{d}{dx} \left(3 x\right) = {\color{red}\left(- \sin{\left(u \right)}\right)} \frac{d}{dx} \left(3 x\right)$$Return to the old variable:
$$- \sin{\left({\color{red}\left(u\right)} \right)} \frac{d}{dx} \left(3 x\right) = - \sin{\left({\color{red}\left(3 x\right)} \right)} \frac{d}{dx} \left(3 x\right)$$Apply the constant multiple rule $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$ with $$$c = 3$$$ and $$$f{\left(x \right)} = x$$$:
$$- \sin{\left(3 x \right)} {\color{red}\left(\frac{d}{dx} \left(3 x\right)\right)} = - \sin{\left(3 x \right)} {\color{red}\left(3 \frac{d}{dx} \left(x\right)\right)}$$Apply the power rule $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ with $$$n = 1$$$, in other words, $$$\frac{d}{dx} \left(x\right) = 1$$$:
$$- 3 \sin{\left(3 x \right)} {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} = - 3 \sin{\left(3 x \right)} {\color{red}\left(1\right)}$$Thus, $$$\frac{d}{dx} \left(\cos{\left(3 x \right)}\right) = - 3 \sin{\left(3 x \right)}$$$.
Answer
$$$\frac{d}{dx} \left(\cos{\left(3 x \right)}\right) = - 3 \sin{\left(3 x \right)}$$$A