derivate ln(1/sqrt(x^2+y^2)) in relation to x

We are tasked with finding the derivative of \( \ln\left( \frac{1}{\sqrt{x^2 + y^2}} \right) \) with respect to \( x \).

### Step-by-Step Solution:

1. **Rewrite the expression:**

   The given expression is:
   \[
   \ln\left( \frac{1}{\sqrt{x^2 + y^2}} \right)
   \]
   We can simplify this as:
   \[
   \ln\left( \frac{1}{\sqrt{x^2 + y^2}} \right) = \ln(1) - \ln(\sqrt{x^2 + y^2})
   \]
   Since \( \ln(1) = 0 \), this simplifies further to:
   \[
   - \ln(\sqrt{x^2 + y^2})
   \]
   Now, use the property of logarithms \( \ln(a^b) = b \ln(a) \) to get:
   \[
   -\frac{1}{2} \ln(x^2 + y^2)
   \]

2. **Differentiate with respect to \( x \):**

   Now, take the derivative of \( -\frac{1}{2} \ln(x^2 + y^2) \) with respect to \( x \). The chain rule gives us:
   \[
   \frac{d}{dx} \left( -\frac{1}{2} \ln(x^2 + y^2) \right) = -\frac{1}{2} \cdot \frac{1}{x^2 + y^2} \cdot \frac{d}{dx}(x^2 + y^2)
   \]

3. **Differentiate \( x^2 + y^2 \):**

   The derivative of \( x^2 + y^2 \) with respect to \( x \) is:
   \[
   \frac{d}{dx}(x^2 + y^2) = 2x
   \]

4. **Substitute back:**

   Substituting this result back into the previous expression:
   \[
   -\frac{1}{2} \cdot \frac{1}{x^2 + y^2} \cdot 2x = \frac{-x}{x^2 + y^2}
   \]

### Final Answer:
The derivative of \( \ln\left( \frac{1}{\sqrt{x^2 + y^2}} \right) \) with respect to \( x \) is:
\[
\frac{-x}{x^2 + y^2}
\]

Would you like further details on any step, or have any other questions?

---

Here are 5 related questions you might find useful:
1. What is the derivative of \( \ln(x^2 + y^2) \) with respect to \( y \)?
2. How do you differentiate \( \sqrt{x^2 + y^2} \) with respect to \( x \)?
3. What is the gradient of the function \( f(x, y) = \ln(x^2 + y^2) \)?
4. How would the result change if the expression inside the logarithm was \( \frac{1}{x^2 + y^2} \)?
5. What is the second derivative of \( \ln\left( \frac{1}{\sqrt{x^2 + y^2}} \right) \) with respect to \( x \)?

**Tip:** For functions of the form \( \ln(f(x)) \), the chain rule always applies by differentiating \( f(x) \) first and multiplying it by \( \frac{1}{f(x)} \).

Learn how to differentiate ln(1/sqrt(x^2 + y^2)) with respect to x using logarithmic properties and the chain rule. Follow a step-by-step solution, including key formulas and concepts, suitable for Grades 11-12 or early college 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