Second derivative of $$$\sin{\left(2 x \right)}$$$
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Find $$$\frac{d^{2}}{dx^{2}} \left(\sin{\left(2 x \right)}\right)$$$.
Solution
Find the first derivative $$$\frac{d}{dx} \left(\sin{\left(2 x \right)}\right)$$$
The function $$$\sin{\left(2 x \right)}$$$ is the composition $$$f{\left(g{\left(x \right)} \right)}$$$ of two functions $$$f{\left(u \right)} = \sin{\left(u \right)}$$$ and $$$g{\left(x \right)} = 2 x$$$.
Apply the chain rule $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:
$${\color{red}\left(\frac{d}{dx} \left(\sin{\left(2 x \right)}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\sin{\left(u \right)}\right) \frac{d}{dx} \left(2 x\right)\right)}$$The derivative of the sine is $$$\frac{d}{du} \left(\sin{\left(u \right)}\right) = \cos{\left(u \right)}$$$:
$${\color{red}\left(\frac{d}{du} \left(\sin{\left(u \right)}\right)\right)} \frac{d}{dx} \left(2 x\right) = {\color{red}\left(\cos{\left(u \right)}\right)} \frac{d}{dx} \left(2 x\right)$$Return to the old variable:
$$\cos{\left({\color{red}\left(u\right)} \right)} \frac{d}{dx} \left(2 x\right) = \cos{\left({\color{red}\left(2 x\right)} \right)} \frac{d}{dx} \left(2 x\right)$$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 $$$c = 2$$$ and $$$f{\left(x \right)} = x$$$:
$$\cos{\left(2 x \right)} {\color{red}\left(\frac{d}{dx} \left(2 x\right)\right)} = \cos{\left(2 x \right)} {\color{red}\left(2 \frac{d}{dx} \left(x\right)\right)}$$Apply the power rule $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ with $$$n = 1$$$, in other words, $$$\frac{d}{dx} \left(x\right) = 1$$$:
$$2 \cos{\left(2 x \right)} {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} = 2 \cos{\left(2 x \right)} {\color{red}\left(1\right)}$$Thus, $$$\frac{d}{dx} \left(\sin{\left(2 x \right)}\right) = 2 \cos{\left(2 x \right)}$$$.
Next, $$$\frac{d^{2}}{dx^{2}} \left(\sin{\left(2 x \right)}\right) = \frac{d}{dx} \left(2 \cos{\left(2 x \right)}\right)$$$
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 $$$c = 2$$$ and $$$f{\left(x \right)} = \cos{\left(2 x \right)}$$$:
$${\color{red}\left(\frac{d}{dx} \left(2 \cos{\left(2 x \right)}\right)\right)} = {\color{red}\left(2 \frac{d}{dx} \left(\cos{\left(2 x \right)}\right)\right)}$$The function $$$\cos{\left(2 x \right)}$$$ is the composition $$$f{\left(g{\left(x \right)} \right)}$$$ of two functions $$$f{\left(u \right)} = \cos{\left(u \right)}$$$ and $$$g{\left(x \right)} = 2 x$$$.
Apply the chain rule $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:
$$2 {\color{red}\left(\frac{d}{dx} \left(\cos{\left(2 x \right)}\right)\right)} = 2 {\color{red}\left(\frac{d}{du} \left(\cos{\left(u \right)}\right) \frac{d}{dx} \left(2 x\right)\right)}$$The derivative of the cosine is $$$\frac{d}{du} \left(\cos{\left(u \right)}\right) = - \sin{\left(u \right)}$$$:
$$2 {\color{red}\left(\frac{d}{du} \left(\cos{\left(u \right)}\right)\right)} \frac{d}{dx} \left(2 x\right) = 2 {\color{red}\left(- \sin{\left(u \right)}\right)} \frac{d}{dx} \left(2 x\right)$$Return to the old variable:
$$- 2 \sin{\left({\color{red}\left(u\right)} \right)} \frac{d}{dx} \left(2 x\right) = - 2 \sin{\left({\color{red}\left(2 x\right)} \right)} \frac{d}{dx} \left(2 x\right)$$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 $$$c = 2$$$ and $$$f{\left(x \right)} = x$$$:
$$- 2 \sin{\left(2 x \right)} {\color{red}\left(\frac{d}{dx} \left(2 x\right)\right)} = - 2 \sin{\left(2 x \right)} {\color{red}\left(2 \frac{d}{dx} \left(x\right)\right)}$$Apply the power rule $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ with $$$n = 1$$$, in other words, $$$\frac{d}{dx} \left(x\right) = 1$$$:
$$- 4 \sin{\left(2 x \right)} {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} = - 4 \sin{\left(2 x \right)} {\color{red}\left(1\right)}$$Thus, $$$\frac{d}{dx} \left(2 \cos{\left(2 x \right)}\right) = - 4 \sin{\left(2 x \right)}$$$.
Therefore, $$$\frac{d^{2}}{dx^{2}} \left(\sin{\left(2 x \right)}\right) = - 4 \sin{\left(2 x \right)}$$$.
Answer
$$$\frac{d^{2}}{dx^{2}} \left(\sin{\left(2 x \right)}\right) = - 4 \sin{\left(2 x \right)}$$$A