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