If u = x^3 - y^3√x + √y, find x di dx + y di dy.

Steps to Solve

1. Differentiate with respect to ( x )

To find (\frac{du}{dx}), we will differentiate each term of the expression ( u = x^3 - y^3 \sqrt{x} + \sqrt{y} ) with respect to ( x ). Note that when differentiating terms that involve ( y ), treat ( y ) as a constant.

The differentiation gives: $$ \frac{du}{dx} = \frac{d}{dx}(x^3) - \frac{d}{dx}(y^3\sqrt{x}) + \frac{d}{dx}(\sqrt{y}) $$

Calculating each part:

• The derivative of ( x^3 ) is ( 3x^2 ).
• The derivative of ( y^3\sqrt{x} ) (using the product rule) is ( y^3 \cdot \frac{1}{2\sqrt{x}} ).
• The derivative of ( \sqrt{y} ) with respect to ( x ) is 0, as ( y ) is constant with respect to ( x ).

So we have: $$ \frac{du}{dx} = 3x^2 - \frac{y^3}{2\sqrt{x}} $$

2. Differentiate with respect to ( y )

Next, we differentiate ( u ) with respect to ( y ) to find (\frac{du}{dy}). Here, treat ( x ) as a constant.

Using the same expression: $$ \frac{du}{dy} = \frac{d}{dy}(x^3) - \frac{d}{dy}(y^3\sqrt{x}) + \frac{d}{dy}(\sqrt{y}) $$

Calculating each part:

• The derivative of ( x^3 ) is 0 (as ( x ) is constant).
• The derivative of ( y^3\sqrt{x} ) is ( 3y^2\sqrt{x} ).
• The derivative of ( \sqrt{y} ) is ( \frac{1}{2\sqrt{y}} ).

Thus we get: $$ \frac{du}{dy} = -3y^2\sqrt{x} + \frac{1}{2\sqrt{y}} $$

3. Summary of derivatives

So, we summarize our findings for the derivatives:

• The result of differentiating with respect to ( x ) is: $$ \frac{du}{dx} = 3x^2 - \frac{y^3}{2\sqrt{x}} $$

• The result of differentiating with respect to ( y ) is: $$ \frac{du}{dy} = -3y^2\sqrt{x} + \frac{1}{2\sqrt{y}} $$