Calculate the first and second derivatives of 𝑔(𝑥) = cos (ln(𝑥" + 1)).

To find the first and second derivatives of the function \( g(x) = \cos(\ln(x^2 + 1)) \), let's break it down step by step.

### 1. First Derivative: \( g'(x) \)

We apply the chain rule for differentiation:

- The outer function is \( \cos(u) \), where \( u = \ln(x^2 + 1) \).
- The derivative of \( \cos(u) \) is \( -\sin(u) \), so:

\[
g'(x) = -\sin(\ln(x^2 + 1)) \cdot \frac{d}{dx} [\ln(x^2 + 1)].
\]

Now, we need to find the derivative of \( \ln(x^2 + 1) \). Using the chain rule again:

- The derivative of \( \ln(v) \), where \( v = x^2 + 1 \), is \( \frac{1}{v} \cdot \frac{d}{dx} [x^2 + 1] \).
- The derivative of \( x^2 + 1 \) is \( 2x \), so:

\[
\frac{d}{dx} [\ln(x^2 + 1)] = \frac{2x}{x^2 + 1}.
\]

Therefore, the first derivative becomes:

\[
g'(x) = -\sin(\ln(x^2 + 1)) \cdot \frac{2x}{x^2 + 1}.
\]

### 2. Second Derivative: \( g''(x) \)

Now, to find the second derivative, we differentiate \( g'(x) = -\sin(\ln(x^2 + 1)) \cdot \frac{2x}{x^2 + 1} \) using the product rule. Let:

- \( f(x) = \sin(\ln(x^2 + 1)) \)
- \( h(x) = \frac{2x}{x^2 + 1} \)

The product rule states:

\[
g''(x) = f'(x) \cdot h(x) + f(x) \cdot h'(x).
\]

#### 2.1. Finding \( f'(x) \):

We already know that \( f(x) = \sin(\ln(x^2 + 1)) \). Using the chain rule:

\[
f'(x) = \cos(\ln(x^2 + 1)) \cdot \frac{d}{dx} [\ln(x^2 + 1)] = \cos(\ln(x^2 + 1)) \cdot \frac{2x}{x^2 + 1}.
\]

#### 2.2. Finding \( h'(x) \):

Now we differentiate \( h(x) = \frac{2x}{x^2 + 1} \) using the quotient rule. The quotient rule is given by:

\[
\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2}.
\]

Here, \( u = 2x \) and \( v = x^2 + 1 \). So:

- \( u' = 2 \),
- \( v' = 2x \),
  
Using the quotient rule:

\[
h'(x) = \frac{(2)(x^2 + 1) - (2x)(2x)}{(x^2 + 1)^2} = \frac{2(x^2 + 1) - 4x^2}{(x^2 + 1)^2} = \frac{2 - 2x^2}{(x^2 + 1)^2}.
\]

#### 2.3. Putting it all together:

Now we can combine the terms for \( g''(x) \):

\[
g''(x) = \cos(\ln(x^2 + 1)) \cdot \frac{2x}{x^2 + 1} \cdot \frac{2x}{x^2 + 1} + \sin(\ln(x^2 + 1)) \cdot \frac{2 - 2x^2}{(x^2 + 1)^2}.
\]

Simplifying each term:

\[
g''(x) = \cos(\ln(x^2 + 1)) \cdot \frac{4x^2}{(x^2 + 1)^2} + \sin(\ln(x^2 + 1)) \cdot \frac{2(1 - x^2)}{(x^2 + 1)^2}.
\]

Thus, the second derivative is:

\[
g''(x) = \frac{4x^2 \cos(\ln(x^2 + 1)) + 2(1 - x^2) \sin(\ln(x^2 + 1))}{(x^2 + 1)^2}.
\]

### Conclusion:

- First derivative: 
  \[
  g'(x) = -\sin(\ln(x^2 + 1)) \cdot \frac{2x}{x^2 + 1}.
  \]
  
- Second derivative: 
  \[
  g''(x) = \frac{4x^2 \cos(\ln(x^2 + 1)) + 2(1 - x^2) \sin(\ln(x^2 + 1))}{(x^2 + 1)^2}.
  \]

Would you like more details on any part of this process?

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Here are 5 related questions to deepen your understanding:

1. How does the chain rule apply to composite functions?
2. What is the geometric interpretation of the first and second derivatives?
3. How do you use the quotient rule when differentiating rational functions?
4. What are some common mistakes made when applying the product rule?
5. How do higher-order derivatives relate to the curvature of a graph?

**Tip:** Always simplify your expression as much as possible after each step to make future calculations easier.

Calculate the first and second derivatives of 𝑔(𝑥) = cos(ln(𝑥^2 + 1)).

Explore the step-by-step process of calculating the first and second derivatives of the function g(x) = cos(ln(x^2 + 1)). This problem involves advanced differentiation techniques such as the chain rule, product rule, and quotient rule, making it suitable for undergraduate calculus students or advanced high school learners.

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