https://file.aiquickdraw.com/mathbot/file/1725590636539o48c0gvq.jpg

The image presents the following problem:

Given the following:
- \( y = \tan u \)
- \( u = \frac{1}{v} \)
- \( v = \ln x \)

You are asked to find the value of \( \frac{dy}{dx} \) at \( x = e^2 \).

### Solution:

We need to find \( \frac{dy}{dx} \) using the chain rule:
\[
\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx}
\]

Let's calculate each part step by step:

1. **Find \( \frac{dy}{du} \):**
   Since \( y = \tan u \),
   \[
   \frac{dy}{du} = \sec^2 u
   \]

2. **Find \( \frac{du}{dv} \):**
   Since \( u = \frac{1}{v} \),
   \[
   \frac{du}{dv} = -\frac{1}{v^2}
   \]

3. **Find \( \frac{dv}{dx} \):**
   Since \( v = \ln x \),
   \[
   \frac{dv}{dx} = \frac{1}{x}
   \]

Now, combining everything:
\[
\frac{dy}{dx} = \sec^2 u \cdot \left( -\frac{1}{v^2} \right) \cdot \frac{1}{x}
\]

### Substitute the expressions for \( u \) and \( v \):
- \( u = \frac{1}{v} = \frac{1}{\ln x} \)
- \( v = \ln x \)

At \( x = e^2 \), we have:


If y = tan(u), u = 1/v, and v = ln(x), what is the value of dy/dx at x = e^2?

Learn how to apply the chain rule to differentiate a composite function where y = tan(u), u = 1/v, and v = ln(x). Find the derivative dy/dx at x = e^2 with step-by-step instructions, including differentiation of trigonometric and logarithmic functions. Suitable for undergraduate 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