Second derivative of $$$x^{e}$$$
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Find $$$\frac{d^{2}}{dx^{2}} \left(x^{e}\right)$$$.
Solution
Find the first derivative $$$\frac{d}{dx} \left(x^{e}\right)$$$
Apply the power rule $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ with $$$n = e$$$:
$${\color{red}\left(\frac{d}{dx} \left(x^{e}\right)\right)} = {\color{red}\left(e x^{-1 + e}\right)}$$Thus, $$$\frac{d}{dx} \left(x^{e}\right) = e x^{-1 + e}$$$.
Next, $$$\frac{d^{2}}{dx^{2}} \left(x^{e}\right) = \frac{d}{dx} \left(e x^{-1 + e}\right)$$$
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 $$$c = e$$$ and $$$f{\left(x \right)} = x^{-1 + e}$$$:
$${\color{red}\left(\frac{d}{dx} \left(e x^{-1 + e}\right)\right)} = {\color{red}\left(e \frac{d}{dx} \left(x^{-1 + e}\right)\right)}$$Apply the power rule $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ with $$$n = -1 + e$$$:
$$e {\color{red}\left(\frac{d}{dx} \left(x^{-1 + e}\right)\right)} = e {\color{red}\left(\left(-1 + e\right) x^{-2 + e}\right)}$$Thus, $$$\frac{d}{dx} \left(e x^{-1 + e}\right) = e x^{-2 + e} \left(-1 + e\right)$$$.
Therefore, $$$\frac{d^{2}}{dx^{2}} \left(x^{e}\right) = e x^{-2 + e} \left(-1 + e\right)$$$.
Answer
$$$\frac{d^{2}}{dx^{2}} \left(x^{e}\right) = e x^{-2 + e} \left(-1 + e\right)$$$A