u=cosecinverse (x^1/2 + y^1/2/ x^1/3 + y^1/3)^1/2

Find du/dx

Given the function \( u = \csc^{-1} \left( \left(\frac{\sqrt{x} + \sqrt{y}}{\sqrt[3]{x} + \sqrt[3]{y}}\right)^{1/2} \right) \), we want to find \(\frac{du}{dx}\).

### Step 1: Simplify the Function
Let:
\[
z = \left(\frac{\sqrt{x} + \sqrt{y}}{\sqrt[3]{x} + \sqrt[3]{y}}\right)^{1/2}
\]
Then, the function becomes:
\[
u = \csc^{-1}(z)
\]

### Step 2: Differentiate \( u \) with respect to \( z \)
Using the chain rule, differentiate \( u \) with respect to \( z \):
\[
\frac{du}{dz} = -\frac{1}{z \sqrt{z^2 - 1}}
\]

### Step 3: Differentiate \( z \) with respect to \( x \)
Now, differentiate \( z \) with respect to \( x \). Start with:
\[
z = \left(\frac{\sqrt{x} + \sqrt{y}}{\sqrt[3]{x} + \sqrt[3]{y}}\right)^{1/2}
\]
Let:
\[
w = \frac{\sqrt{x} + \sqrt{y}}{\sqrt[3]{x} + \sqrt[3]{y}}
\]
Then:
\[
z = w^{1/2}
\]
So:
\[
\frac{dz}{dx} = \frac{1}{2} w^{-1/2} \cdot \frac{dw}{dx}
\]

### Step 4: Differentiate \( w \) with respect to \( x \)
Now, differentiate \( w \) with respect to \( x \):
\[
w = \frac{\sqrt{x} + \sqrt{y}}{\sqrt[3]{x} + \sqrt[3]{y}}
\]
Using the quotient rule:
\[
\frac{dw}{dx} = \frac{\left(\frac{1}{2}x^{-1/2}\right)(\sqrt[3]{x} + \sqrt[3]{y}) - (\sqrt{x} + \sqrt{y})\left(\frac{1}{3}x^{-2/3}\right)}{(\sqrt[3]{x} + \sqrt[3]{y})^2}
\]

### Step 5: Combine Results
Finally, combine all these steps to find \(\frac{du}{dx}\):
\[
\frac{du}{dx} = \frac{du}{dz} \cdot \frac{dz}{dx}
\]

Substitute \( z \), \( \frac{du}{dz} \), \( \frac{dz}{dx} \), and \( \frac{dw}{dx} \):
\[
\frac{du}{dx} = -\frac{1}{z \sqrt{z^2 - 1}} \cdot \frac{1}{2} w^{-1/2} \cdot \frac{\frac{1}{2}x^{-1/2}(\sqrt[3]{x} + \sqrt[3]{y}) - (\sqrt{x} + \sqrt{y})\frac{1}{3}x^{-2/3}}{(\sqrt[3]{x} + \sqrt[3]{y})^2}
\]
Where:
\[
w = \frac{\sqrt{x} + \sqrt{y}}{\sqrt[3]{x} + \sqrt[3]{y}}, \quad z = \sqrt{w}
\]

This is the expression for \(\frac{du}{dx}\). It represents the derivative of the given function with respect to \( x \).

Would you like more details or have any specific questions?

### Related Questions
1. How does the derivative \(\frac{du}{dx}\) change when \( y \) is held constant?
2. Can you evaluate \(\frac{du}{dx}\) for specific values of \( x \) and \( y \)?
3. What is the interpretation of this derivative in a physical context?
4. How would \(\frac{du}{dx}\) change if the trigonometric function were \( \sec^{-1} \) instead of \( \csc^{-1} \)?
5. Can you simplify the expression for small or large values of \( x \)?
6. How does \(\frac{du}{dx}\) behave as \( x \) approaches zero?
7. What is the impact of changing the powers \( \frac{1}{2} \) and \( \frac{1}{3} \) on the derivative?
8. How does the expression for \(\frac{du}{dx}\) compare to \(\frac{du}{dy}\)?

### Tip
When faced with complex derivatives, breaking down the expression into simpler parts and using substitution can make the differentiation process more manageable.

Explore how to differentiate the inverse cosecant function involving complex trigonometric ratios. Learn step-by-step with detailed solutions using differentiation rules. Suitable for advanced high school students and college-level learners.

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