Let $f(x)=\sqrt{x^2+8}$.
Find:
\(f'(x) = \frac{x}{\sqrt{x^2 + 8}}\), \(f''(x) = \frac{8}{(x^2 + 8)^{3/2}}\), and \(f''(3) = \boxed{\frac{8\sqrt{17}}{289}}\).
The given function is \(f(x) = \sqrt{x^2 + 8}\). We can express this function using a power notation as \(f(x) = (x^2 + 8)^{1/2}\). This form is convenient for applying the chain rule, which states that for a function of the form \(h(x) = g(u(x))\), the derivative is \(h'(x) = g'(u(x)) \cdot u'(x)\). Here, \(g(u) = u^{1/2}\) and \(u(x) = x^2 + 8\). The derivative of \(g(u)\) is \(g'(u) = \frac{1}{2}u^{-1/2}\), and the derivative of \(u(x)\) is \(u'(x) = 2x\). Combining these, we get:
\[f'(x) = \frac{1}{2}(x^2 + 8)^{-1/2} \cdot (2x) = x(x^2 + 8)^{-1/2} = \frac{x}{\sqrt{x^2 + 8}}\]
To find the second derivative \(f''(x)\), we differentiate \(f'(x) = \frac{x}{\sqrt{x^2 + 8}}\) with respect to \(x\) using the quotient rule \(\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}\), where \(u = x\) and \(v = \sqrt{x^2 + 8}\).
We know that \(u' = 1\) and \(v' = \frac{d}{dx}((x^2+8)^{1/2}) = \frac{x}{\sqrt{x^2 + 8}}\).
Applying the rule:
\[f''(x) = \frac{1 \cdot \sqrt{x^2 + 8} - x \cdot \left(\frac{x}{\sqrt{x^2 + 8}}\right)}{(\sqrt{x^2 + 8})^2}\]
To simplify, multiply the numerator and denominator by \(\sqrt{x^2 + 8}\):
\[f''(x) = \frac{(x^2 + 8) - x^2}{(x^2 + 8)\sqrt{x^2 + 8}} = \frac{8}{(x^2 + 8)^{3/2}}\]
Now we substitute \(x = 3\) into the expression for the second derivative:
\[f''(3) = \frac{8}{(3^2 + 8)^{3/2}} = \frac{8}{(9 + 8)^{3/2}} = \frac{8}{17^{3/2}}\]
Since \(17^{3/2} = 17\sqrt{17}\), we can rewrite this as:
\[f''(3) = \frac{8}{17\sqrt{17}}\]
To rationalize the denominator, multiply by \(\frac{\sqrt{17}}{\sqrt{17}}\):
\[f''(3) = \frac{8\sqrt{17}}{17 \cdot 17} = \frac{8\sqrt{17}}{289}\]
The results are:
\[f'(x) = \frac{x}{\sqrt{x^2 + 8}}\]
\[f''(x) = \frac{8}{(x^2 + 8)^{3/2}}\]
\[f''(3) = \frac{8\sqrt{17}}{289}\]
In physics, if the position of a particle moving along a line is given by \(f(x) = \sqrt{x^2 + 8}\), where \(x\) represents time, the first derivative \(f'(x)\) gives the velocity of the particle, and the second derivative \(f''(x)\) provides the acceleration. By calculating the second derivative, we can determine the force acting on the particle at any given time, such as at \(x=3\), which is crucial in fields like classical mechanics to predict motion and ensure structural stability in various engineering designs.