Find first order partial derivative of u=x^xy

To find the first-order partial derivatives of \( u = x^{x^y} \) with respect to \( x \) and \( y \), we'll proceed step by step.

### 1. Partial derivative of \( u \) with respect to \( x \):

We have:
\[
u = x^{x^y}
\]
First, rewrite \( u \) in logarithmic form to simplify the differentiation process:
\[
u = e^{\ln(x^{x^y})} = e^{x^y \ln x}
\]
Now differentiate \( u \) with respect to \( x \) using the chain rule.

\[
\frac{\partial u}{\partial x} = e^{x^y \ln x} \cdot \frac{d}{dx}(x^y \ln x)
\]
We will apply the product rule to differentiate \( x^y \ln x \):
\[
\frac{d}{dx}(x^y \ln x) = yx^{y-1} \ln x + \frac{x^y}{x} = yx^{y-1} \ln x + x^{y-1}
\]
Thus, the partial derivative with respect to \( x \) is:
\[
\frac{\partial u}{\partial x} = x^{x^y} \left( yx^{y-1} \ln x + x^{y-1} \right)
\]

### 2. Partial derivative of \( u \) with respect to \( y \):

We start again with:
\[
u = x^{x^y} = e^{x^y \ln x}
\]
Now differentiate \( u \) with respect to \( y \):
\[
\frac{\partial u}{\partial y} = e^{x^y \ln x} \cdot \frac{d}{dy}(x^y \ln x)
\]
The derivative of \( x^y \ln x \) with respect to \( y \) is:
\[
\frac{d}{dy}(x^y \ln x) = x^y \ln x \cdot \ln x = x^y (\ln x)^2
\]
Thus, the partial derivative with respect to \( y \) is:
\[
\frac{\partial u}{\partial y} = x^{x^y} (\ln x)^2
\]

### Final Results:

- The partial derivative of \( u \) with respect to \( x \):
  \[
  \frac{\partial u}{\partial x} = x^{x^y} \left( yx^{y-1} \ln x + x^{y-1} \right)
  \]

- The partial derivative of \( u \) with respect to \( y \):
  \[
  \frac{\partial u}{\partial y} = x^{x^y} (\ln x)^2
  \]

Would you like more details on these steps or further clarifications?

### Related Questions:
1. How would you find the second-order partial derivatives of \( u = x^{x^y} \)?
2. What are some applications of partial derivatives in physics and economics?
3. How does the chain rule work when differentiating implicit functions?
4. Can you explain the significance of the product rule in partial differentiation?
5. How would you use the gradient of \( u = x^{x^y} \) in a real-world scenario?

**Tip**: Always remember that taking the natural logarithm can simplify the differentiation of functions involving powers and exponents.

Find first order partial derivative of u = x^x^y

This page provides a detailed solution for finding the first-order partial derivatives of u = x^x^y with respect to x and y. Learn step-by-step methods using the chain rule and product rule, 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