Second derivative of a square-root function

Write $f(x) = (x^2+9)^{1/2}$. By the chain rule, $f'(x) = \frac{1}{2}(x^2+9)^{-1/2}\cdot 2x = \frac{x}{\sqrt{x^2+9}}$. This is the first derivative, expressed as a quotient of $x$ over $\sqrt{x^2+9}$.

Now apply the quotient rule to $f'(x) = \dfrac{x}{\sqrt{x^2+9}}$: $f''(x) = \dfrac{1\cdot\sqrt{x^2+9} - x\cdot \frac{x}{\sqrt{x^2+9}}}{x^2+9}$. Multiply numerator and denominator by $\sqrt{x^2+9}$ to clear the inner fraction, giving numerator $(x^2+9) - x^2 = 9$ and denominator $(x^2+9)^{3/2}$. So $f''(x) = \dfrac{9}{(x^2+9)^{3/2}}$.