Second derivative of $$$\sqrt{x}$$$
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Find $$$\frac{d^{2}}{dx^{2}} \left(\sqrt{x}\right)$$$.
Solution
Find the first derivative $$$\frac{d}{dx} \left(\sqrt{x}\right)$$$
Apply the power rule $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ with $$$n = \frac{1}{2}$$$:
$${\color{red}\left(\frac{d}{dx} \left(\sqrt{x}\right)\right)} = {\color{red}\left(\frac{1}{2 \sqrt{x}}\right)}$$Thus, $$$\frac{d}{dx} \left(\sqrt{x}\right) = \frac{1}{2 \sqrt{x}}$$$.
Next, $$$\frac{d^{2}}{dx^{2}} \left(\sqrt{x}\right) = \frac{d}{dx} \left(\frac{1}{2 \sqrt{x}}\right)$$$
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 $$$c = \frac{1}{2}$$$ and $$$f{\left(x \right)} = \frac{1}{\sqrt{x}}$$$:
$${\color{red}\left(\frac{d}{dx} \left(\frac{1}{2 \sqrt{x}}\right)\right)} = {\color{red}\left(\frac{\frac{d}{dx} \left(\frac{1}{\sqrt{x}}\right)}{2}\right)}$$Apply the power rule $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ with $$$n = - \frac{1}{2}$$$:
$$\frac{{\color{red}\left(\frac{d}{dx} \left(\frac{1}{\sqrt{x}}\right)\right)}}{2} = \frac{{\color{red}\left(- \frac{1}{2 x^{\frac{3}{2}}}\right)}}{2}$$Thus, $$$\frac{d}{dx} \left(\frac{1}{2 \sqrt{x}}\right) = - \frac{1}{4 x^{\frac{3}{2}}}$$$.
Therefore, $$$\frac{d^{2}}{dx^{2}} \left(\sqrt{x}\right) = - \frac{1}{4 x^{\frac{3}{2}}}$$$.
Answer
$$$\frac{d^{2}}{dx^{2}} \left(\sqrt{x}\right) = - \frac{1}{4 x^{\frac{3}{2}}}$$$A