attempt 1: 10 attempts remaining. let $f(x)=sqrt{x^{2}+9}$. $f(x)=$ $f(x)=$ $f(3)=$ submit answer next item

Explanation:

Step1: Apply chain - rule for first - derivative

Let $u = x^{2}+9$, then $y=\sqrt{u}=u^{\frac{1}{2}}$. The chain - rule states that $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. We know that $\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}$ and $\frac{du}{dx} = 2x$. Substituting $u=x^{2}+9$ back in, we get $f'(x)=\frac{2x}{2\sqrt{x^{2}+9}}=\frac{x}{\sqrt{x^{2}+9}}$.

Step2: Apply quotient - rule for second - derivative

The quotient - rule is $\left(\frac{g(x)}{h(x)}\right)'=\frac{g'(x)h(x)-g(x)h'(x)}{h^{2}(x)}$. Here, $g(x)=x$, $g'(x) = 1$, $h(x)=\sqrt{x^{2}+9}=(x^{2}+9)^{\frac{1}{2}}$, and $h'(x)=\frac{x}{\sqrt{x^{2}+9}}$ (from Step 1). Then $f''(x)=\frac{\sqrt{x^{2}+9}-x\cdot\frac{x}{\sqrt{x^{2}+9}}}{x^{2}+9}=\frac{x^{2}+9 - x^{2}}{(x^{2}+9)^{\frac{3}{2}}}=\frac{9}{(x^{2}+9)^{\frac{3}{2}}}$.

Step3: Evaluate $f''(3)$

Substitute $x = 3$ into $f''(x)$. We have $f''(3)=\frac{9}{(3^{2}+9)^{\frac{3}{2}}}=\frac{9}{(9 + 9)^{\frac{3}{2}}}=\frac{9}{(18)^{\frac{3}{2}}}=\frac{9}{18\sqrt{18}}=\frac{1}{2\sqrt{18}}=\frac{1}{6\sqrt{2}}=\frac{\sqrt{2}}{12}$.

Answer:

$f'(x)=\frac{x}{\sqrt{x^{2}+9}}$ $f''(x)=\frac{9}{(x^{2}+9)^{\frac{3}{2}}}$ $f''(3)=\frac{\sqrt{2}}{12}$