Differentiate x^(sinx+1) w.r.t logcot inverse x

To differentiate the function \( f(x) = x^{\sin x + 1} \) with respect to \( \log(\cot^{-1}x) \), we will apply the chain rule. Let's break down the process into steps.

### Step 1: Differentiate \( f(x) = x^{\sin x + 1} \) with respect to \( x \)
The derivative of \( f(x) = x^{\sin x + 1} \) with respect to \( x \) involves using the chain rule and the power rule.

First, express \( f(x) \) as:

\[
f(x) = e^{(\sin x + 1) \ln x}
\]

Now, differentiate this expression with respect to \( x \):

\[
\frac{d}{dx} \left( e^{(\sin x + 1) \ln x} \right)
\]

Using the chain rule:

\[
\frac{d}{dx} \left( e^{(\sin x + 1) \ln x} \right) = e^{(\sin x + 1) \ln x} \cdot \frac{d}{dx} \left( (\sin x + 1) \ln x \right)
\]

Now, differentiate \( (\sin x + 1) \ln x \) with respect to \( x \):

\[
\frac{d}{dx} \left( (\sin x + 1) \ln x \right) = (\cos x) \ln x + \frac{\sin x + 1}{x}
\]

Thus, the derivative of \( f(x) \) with respect to \( x \) is:

\[
\frac{d}{dx} \left( x^{\sin x + 1} \right) = x^{\sin x + 1} \cdot \left( \cos x \cdot \ln x + \frac{\sin x + 1}{x} \right)
\]

### Step 2: Differentiate \( g(x) = \log(\cot^{-1}x) \) with respect to \( x \)
The derivative of \( g(x) = \log(\cot^{-1}x) \) with respect to \( x \) is:

\[
\frac{d}{dx} \left( \log(\cot^{-1}x) \right) = \frac{1}{\cot^{-1}x} \cdot \frac{d}{dx} \left( \cot^{-1}x \right)
\]

The derivative of \( \cot^{-1}x \) with respect to \( x \) is:

\[
\frac{d}{dx} \left( \cot^{-1}x \right) = -\frac{1}{1 + x^2}
\]

Thus, the derivative of \( g(x) \) is:

\[
\frac{d}{dx} \left( \log(\cot^{-1}x) \right) = -\frac{1}{\cot^{-1}x} \cdot \frac{1}{1 + x^2}
\]

### Step 3: Apply the Chain Rule
Now, using the chain rule, the derivative of \( f(x) = x^{\sin x + 1} \) with respect to \( \log(\cot^{-1}x) \) is:

\[
\frac{d}{d(\log(\cot^{-1}x))} \left( x^{\sin x + 1} \right) = \frac{\frac{d}{dx} \left( x^{\sin x + 1} \right)}{\frac{d}{dx} \left( \log(\cot^{-1}x) \right)}
\]

Substituting the results from Step 1 and Step 2:

\[
\frac{\frac{d}{dx} \left( x^{\sin x + 1} \right)}{\frac{d}{dx} \left( \log(\cot^{-1}x) \right)} = \frac{x^{\sin x + 1} \cdot \left( \cos x \cdot \ln x + \frac{\sin x + 1}{x} \right)}{-\frac{1}{\cot^{-1}x} \cdot \frac{1}{1 + x^2}}
\]

Simplifying:

\[
\boxed{-x^{\sin x + 1} \cdot \left( \cos x \cdot \ln x + \frac{\sin x + 1}{x} \right) \cdot \frac{\cot^{-1}x}{1 + x^2}}
\]

This is the derivative of the given function with respect to \( \log(\cot^{-1}x) \).

If you need further clarification or have any questions, feel free to ask!

Here are 5 related questions:
1. How would you differentiate \( x^{\tan x + 1} \) with respect to \( \log(\sec^{-1}x) \)?
2. What is the derivative of \( x^{\cos x + 2} \) with respect to \( \arctan x \)?
3. How do you differentiate \( e^{\sin x + 1} \) with respect to \( \log(\csc^{-1}x) \)?
4. Can you find the second derivative of \( x^{\sin x + 1} \) with respect to \( x \)?
5. What is the derivative of \( \sin^{-1}(x^{2}) \) with respect to \( \log(\cot^{-1}x) \)?

**Tip:** For complex expressions involving logarithms and trigonometric functions, using logarithmic differentiation can simplify the process.

Learn how to differentiate x^(sinx+1) with respect to log(\cot^{-1}x) using the chain rule and logarithmic differentiation. This detailed solution covers the steps involved in finding the derivative of a complex exponential function with respect to a logarithmic expression.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation