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The function given in the problem is:

\[
g(x) = \frac{x (\ln(x))^2 - \sqrt{x^2 + 1}}{x^2 + 1}
\]

We are tasked with finding \( g'(x) \), the derivative of this function.

To solve this, we'll apply the **quotient rule**. The quotient rule is given by:

\[
\left( \frac{f(x)}{h(x)} \right)' = \frac{f'(x)h(x) - f(x)h'(x)}{(h(x))^2}
\]

In this case:
- \( f(x) = x (\ln(x))^2 - \sqrt{x^2 + 1} \)
- \( h(x) = x^2 + 1 \)

### Step 1: Derivatives of \( f(x) \) and \( h(x) \)

- \( h'(x) = 2x \)
- Now, let's break down \( f(x) = x (\ln(x))^2 - \sqrt{x^2 + 1} \):

   - For \( x (\ln(x))^2 \), apply the **product rule**:
     \[
     \frac{d}{dx} \left( x (\ln(x))^2 \right) = (\ln(x))^2 + 2 \ln(x)
     \]
     (since the derivative of \( x \) is 1 and the derivative of \( (\ln(x))^2 \) is \( 2\ln(x) \cdot \frac{1}{x} \)).

   - For \( \sqrt{x^2 + 1} \), apply the **chain rule**:
     \[
     \frac{d}{dx} \left( \sqrt{x^2 + 1} \right) = \frac{x}{\sqrt{x^2 + 1}}
     \]

### Step 2: Putting it all together

Now, applying the quotient rule:

\[
g'(x) = \frac{f'(x)(x^2 + 1) - f(x)(2x)}{(x^2 + 1)^2}
\]

Substitute \( f(x) \), \( f'(x) \), and \( h'(x) \) into this equation to get the final expression for \( g'(x) \).

Would you like the expression fully simplified or any further explanation? Here are some related questions you might find helpful:

1. How do you apply the chain rule to more complex functions?
2. What happens when we differentiate logarithmic expressions?
3. Can you explain the product rule in more detail?
4. What is the importance of the quotient rule in calculus?
5. How does simplifying the derivative impact the final result?

**Tip:** Breaking down each derivative step-by-step makes complex derivative calculations easier to handle.

Let g(x) = \frac{x(\ln(x))^2 - \sqrt{x^2 + 1}}{x^2 + 1}. Find g'(x).

This solution shows how to find the derivative of g(x) = (x(ln(x))^2 - sqrt(x^2 + 1)) / (x^2 + 1) using the quotient, product, and chain rules. The problem involves breaking down the expressions and applying calculus rules step-by-step, suitable for students in undergraduate calculus courses.

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