https://file.aiquickdraw.com/mathbot/file/172963203612709d71wcr.jpg

The given problem asks to show that:

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

### Step-by-Step Solution:

#### 1. Let \( f(x) = \ln\left(x + \sqrt{x^2 + 1}\right) \).
We need to find the derivative of \( f(x) \).

#### 2. Use the chain rule for differentiation:
The derivative of \( \ln(u) \) is \( \frac{1}{u} \cdot \frac{du}{dx} \), so applying this to our function:

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

#### 3. Differentiate the inner function \( x + \sqrt{x^2 + 1} \):
- The derivative of \( x \) is \( 1 \).
- The derivative of \( \sqrt{x^2 + 1} \) is:
  \[
  \frac{d}{dx} \left( \sqrt{x^2 + 1} \right) = \frac{1}{2\sqrt{x^2 + 1}} \cdot 2x = \frac{x}{\sqrt{x^2 + 1}}.
  \]

Thus, the derivative of \( x + \sqrt{x^2 + 1} \) is:
\[
1 + \frac{x}{\sqrt{x^2 + 1}}.
\]

#### 4. Substitute back into the derivative:
Now substitute this result into the expression for \( f'(x) \):

\[
f'(x) = \frac{1}{x + \sqrt{x^2 + 1}} \cdot \left( 1 + \frac{x}{\sqrt{x^2 + 1}} \right).
\]

#### 5. Simplify the expression:
Combine the terms inside the parentheses:

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

Get a common denominator for the numerator:

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

Now simplify by canceling the \( x + \sqrt{x^2 + 1} \) terms:

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

This shows that:

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

Thus, the required result is proved.

---

Would you like any further clarification or more details on any step?

Here are some related questions for further practice:
1. What is the derivative of \( \ln(x + \sqrt{x^2 - 1}) \)?
2. How does the chain rule apply to functions involving square roots?
3. Can you generalize the derivative of \( \ln(u(x)) \) for any \( u(x) \)?
4. What would the derivative of \( \ln(x^2 + \sqrt{x^4 + 1}) \) be?
5. How would you apply the quotient rule to a problem like this?

**Tip:** When differentiating logarithmic functions, using the chain rule carefully is key to obtaining the correct result, especially when there are nested functions involved.

Show that the derivative of ln(x + √(x^2 + 1)) is equal to 1 / √(x^2 + 1).

Find the derivative of ln(x + √(x^2 + 1)) using the chain rule for differentiation. Step-by-step solution, suitable for university-level calculus students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation