Problem 24 - Calculus Early Transcendentals

Question

Find the first partial derivatives of the following functions. $$f(x, y)=\sqrt{x^{2} y^{3}}$$

Step-by-Step Solution

Answer

Answer: The first partial derivatives of the function are: $$\frac{\partial f}{\partial x} = \frac{y}{\sqrt{x}}$$ $$\frac{\partial f}{\partial y} = \frac{3x}{2\sqrt{y}}$$

Step 1: Rewrite function as a power

Rewrite the function in power notation, it will make differentiation easier: $$f(x, y) = (x^{2} y^{3})^{\frac{1}{2}}$$

Step 2: Differentiate with respect to x

Using the chain rule: $$\frac{\partial f}{\partial x} = \frac{1}{2}(x^{2} y^{3})^{-\frac{1}{2}} \cdot \frac{\partial}{\partial x} (x^{2} y^{3})$$ Now we need to differentiate the term inside the parentheses with respect to x: $$\frac{\partial}{\partial x}(x^{2} y^{3}) = 2x y^{3}$$

Step 3: Substitute back into the formula

Substitute the differentiated term back into the formula for the partial derivative: $$\frac{\partial f}{\partial x} = \frac{1}{2}(x^{2} y^{3})^{-\frac{1}{2}} \cdot (2xy^{3})$$

Step 4: Simplify the expression

Simplify the expression to obtain our final result: $$\frac{\partial f}{\partial x} = \frac{2xy^{3}}{2\sqrt{x^{2}y^{3}}} = \frac{y}{\sqrt{x}}$$ Next, we will find the partial derivative of f with respect to y, denoted as \(\frac{\partial f}{\partial y}\).

Step 5: Differentiate with respect to y

Using the chain rule again: $$\frac{\partial f}{\partial y} = \frac{1}{2}(x^{2} y^{3})^{-\frac{1}{2}} \cdot \frac{\partial}{\partial y} (x^{2} y^{3})$$ We need to differentiate the term inside the parentheses with respect to y: $$\frac{\partial}{\partial y}(x^{2} y^{3}) = 3x^{2} y^{2}$$

Step 6: Substitute back into the formula

Substitute the differentiated term back into the formula for the partial derivative: $$\frac{\partial f}{\partial y} = \frac{1}{2}(x^{2} y^{3})^{-\frac{1}{2}} \cdot (3x^{2}y^{2})$$

Step 7: Simplify the expression

Simplify the expression to obtain our final result: $$\frac{\partial f}{\partial y} = \frac{3x^{2}y^{2}}{2\sqrt{x^{2}y^{3}}} = \frac{3x}{2\sqrt{y}}$$ So, the first partial derivatives of the function are: $$\frac{\partial f}{\partial x} = \frac{y}{\sqrt{x}}$$ $$\frac{\partial f}{\partial y} = \frac{3x}{2\sqrt{y}}$$