Problem 20 - The Calculus with Analytic Geometry

Question

Consider \(y\) as the independent variable and find \(D_{y} x\). $$ y \sqrt{x}-x \sqrt{y}=9 $$

Step-by-Step Solution

Answer

Implicit differentiation gives \(D_y(x) = \frac{ \frac{x}{2\sqrt{y}} - \sqrt{x} }{ \frac{y}{2\sqrt{x}} - \sqrt{y} } \).

Step 1 - Rewrite the equation

Start with the given equation: \[ y \sqrt{x} - x \sqrt{y} = 9 \].

Step 2 - Differentiate both sides with respect to y

Apply implicit differentiation to both sides of the equation with respect to y. Remember that y is the independent variable and x is a function of y: \[ D_y(y \sqrt{x}) - D_y(x \sqrt{y}) = D_y(9) \].

Step 3 - Use the product rule

Use the product rule to differentiate each term. For the left term: \[ D_y(y \sqrt{x}) = y \frac{d}{dy}(\sqrt{x}) + \sqrt{x} \frac{d}{dy}(y) = y \frac{1}{2\sqrt{x}}D_y(x) + \sqrt{x} \]. For the right term: \[ D_y(x \sqrt{y}) = x \frac{d}{dy}(\sqrt{y}) + \sqrt{y} \frac{d}{dy}(x) = x \frac{1}{2\sqrt{y}} + \sqrt{y}D_y(x) \].

Step 4 - Substitute derivatives

Now substitute back the differentiated terms into the original equation: \[ y \frac{1}{2\sqrt{x}}D_y(x) + \sqrt{x} - x \frac{1}{2\sqrt{y}} = 0 \].

Step 5 - Isolate the derivative term

Rearrange the equation to isolate \( D_y(x) \). Combine like terms and solve for the derivative: \[ y \frac{1}{2\sqrt{x}}D_y(x) - \sqrt{y}D_y(x) = x \frac{1}{2\sqrt{y}} - \sqrt{x} \]. Factor out \( D_y(x) \): \[ D_y(x)\left( y \frac{1}{2\sqrt{x}} - \sqrt{y} \right) = x \frac{1}{2\sqrt{y}} - \sqrt{x} \].

Step 6 - Solve for D_y(x)

Finally, solve for \( D_y(x) \) by dividing both sides by the common factor: \[ D_y(x) = \frac{ x \frac{1}{2\sqrt{y}} - \sqrt{x} }{ y \frac{1}{2\sqrt{x}} - \sqrt{y} } \]. Simplify where possible: \[ D_y(x) = \frac{ \frac{x}{2\sqrt{y}} - \sqrt{x} }{ \frac{y}{2\sqrt{x}} - \sqrt{y} } \].

Key Concepts

Derivative

A derivative is a measure of how a function changes as its input changes. It tells us the rate at which one quantity changes with respect to another. For any function, its derivative provides the slope of the tangent line at any point on the graph. In the context of implicit differentiation, we find the derivative with respect to an independent variable, but instead of solving for one variable in terms of another, we differentiate both sides of an equation involving multiple variables. This allows us to handle cases where the dependent variable cannot be easily isolated, as seen in equations like \[ y \sqrt{x} - x \sqrt{y} = 9 \].

Product Rule

The product rule is a fundamental tool in calculus used to differentiate the product of two functions. When we have a product of two functions, say \( u(y) \) and \( v(y) \), its derivative is given by the formula: \( (uv)' = u'v + uv' \). This means we differentiate the first function and multiply it by the second function, then add the result to the first function multiplied by the derivative of the second function.

In implicit differentiation, we often apply the product rule because we deal with products of functions involving both dependent and independent variables. For instance, to differentiate \( y \sqrt{x} \) with respect to \( y \), we use the product rule to get: \[ D_y(y \sqrt{x}) = y \frac{d}{dy}(\sqrt{x}) + \sqrt{x} \frac{d}{dy}(y) = y \frac{1}{2\sqrt{x}}D_y(x) + \sqrt{x} \]. This approach simplifies finding derivatives in complex situations.

Dependent and Independent Variables

In calculus, it’s crucial to distinguish between independent and dependent variables. The independent variable is the one you control or choose values for, while the dependent variable depends on the independent variable's values. For example, in the equation \( y \sqrt{x} - x \sqrt{y} = 9 \), \( y \) is independent, and \( x \) is dependent because \( x \) changes based on \( y \).

When we perform implicit differentiation, we treat the independent variable as a constant when differentiating expressions involving the dependent variable. This allows us to find how the dependent variable's rate of change with respect to the independent variable. The process involves applying differentiation rules like the product rule and carefully handling each variable according to their status as dependent or independent.