homework5: problem 42 (1 point) let $f(x)=sqrt{x^{2}+9}$. $f(x)=$ $f(1)=$ $f(x)=$ $f(1)=$

Explanation:

Step1: Apply chain - rule for first - derivative

Let $u = x^{2}+9$, then $y = \sqrt{u}=u^{\frac{1}{2}}$. By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. We have $\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}$ and $\frac{du}{dx}=2x$. So $f^{\prime}(x)=\frac{2x}{2\sqrt{x^{2}+9}}=\frac{x}{\sqrt{x^{2}+9}}$.

Step2: Evaluate $f^{\prime}(1)$

Substitute $x = 1$ into $f^{\prime}(x)$. Then $f^{\prime}(1)=\frac{1}{\sqrt{1 + 9}}=\frac{1}{\sqrt{10}}$.

Step3: Apply quotient - rule for second - derivative

The quotient - rule states that if $y=\frac{g(x)}{h(x)}$, then $y^{\prime}=\frac{g^{\prime}(x)h(x)-g(x)h^{\prime}(x)}{h^{2}(x)}$. Here, $g(x)=x$, $g^{\prime}(x)=1$, $h(x)=\sqrt{x^{2}+9}=(x^{2}+9)^{\frac{1}{2}}$, and $h^{\prime}(x)=\frac{x}{\sqrt{x^{2}+9}}$. So $f^{\prime\prime}(x)=\frac{\sqrt{x^{2}+9}-x\cdot\frac{x}{\sqrt{x^{2}+9}}}{x^{2}+9}=\frac{\frac{x^{2}+9 - x^{2}}{\sqrt{x^{2}+9}}}{x^{2}+9}=\frac{9}{(x^{2}+9)^{\frac{3}{2}}}$.

Step4: Evaluate $f^{\prime\prime}(1)$

Substitute $x = 1$ into $f^{\prime\prime}(x)$. Then $f^{\prime\prime}(1)=\frac{9}{(1 + 9)^{\frac{3}{2}}}=\frac{9}{10^{\frac{3}{2}}}=\frac{9}{10\sqrt{10}}$.

Answer:

$f^{\prime}(x)=\frac{x}{\sqrt{x^{2}+9}}$ $f^{\prime}(1)=\frac{1}{\sqrt{10}}$ $f^{\prime\prime}(x)=\frac{9}{(x^{2}+9)^{\frac{3}{2}}}$ $f^{\prime\prime}(1)=\frac{9}{10\sqrt{10}}$