Find $$ y^{\prime} $$ and $$ y^{\prime \prime} $$ $$ y=\ln \left(x+\sqrt{9+x^{2}}\right) $$ $$ y^{\prime}=\square $$ $$ y^{\prime \prime}=\square $$

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1. Given the function
   $$
   y = \ln \left(x+\sqrt{9+x^{2}}\right),
   $$
   we differentiate using the chain rule. Note that
   $$
   y^{\prime}=\frac{1}{x+\sqrt{9+x^2}} \cdot \frac{d}{dx}\left(x+\sqrt{9+x^2}\right).
   $$

2. Compute the derivative of the inner function:
   $$
   \frac{d}{dx}\left(x+\sqrt{9+x^2}\right)=1+\frac{1}{2\sqrt{9+x^2}} \cdot 2x=1+\frac{x}{\sqrt{9+x^2}}.
   $$

3. Substitute back into the expression for $$ y^{\prime} $$:
   $$
   y^{\prime}=\frac{1+\frac{x}{\sqrt{9+x^2}}}{x+\sqrt{9+x^2}}.
   $$

4. Simplify the numerator by writing it with a common denominator:
   $$
   1+\frac{x}{\sqrt{9+x^2}}=\frac{\sqrt{9+x^2}+x}{\sqrt{9+x^2}}.
   $$

5. Now substitute:
   $$
   y^{\prime}=\frac{\frac{\sqrt{9+x^2}+x}{\sqrt{9+x^2}}}{x+\sqrt{9+x^2}}=\frac{\sqrt{9+x^2}+x}{\sqrt{9+x^2}\left(x+\sqrt{9+x^2}\right)}.
   $$

6. Notice that the factor $$\sqrt{9+x^2}+x$$ cancels:
   $$
   y^{\prime}=\frac{1}{\sqrt{9+x^2}}.
   $$

7. For the second derivative, we differentiate:
   $$
   y^{\prime}=\left(9+x^2\right)^{-\frac{1}{2}}.
   $$
   Apply the chain rule:
   $$
   y^{\prime\prime}=-\frac{1}{2}\left(9+x^2\right)^{-\frac{3}{2}}\cdot (2x) = -\frac{x}{\left(9+x^2\right)^{\frac{3}{2}}}.
   $$

Thus,
$$
y^{\prime}=\frac{1}{\sqrt{9+x^2}}, \quad y^{\prime\prime}=-\frac{x}{\left(9+x^2\right)^{\frac{3}{2}}}.
$$
$$ y^{\prime} = \frac{1}{\sqrt{9+x^2}} $$ $$ y^{\prime \prime} = -\frac{x}{(9+x^2)^{3/2}} $$

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