Problem 7 - The Calculus with Analytic Geometry

Question

Find \(D_{x} y\) by implicit differentiation. $$ \sqrt{x}+\sqrt{y}=4 $$

Step-by-Step Solution

Answer

\(\frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}}\)

Step 1 - Differentiate both sides with respect to x

Start by differentiating the given equation \ \ \ \( \sqrt{x} + \sqrt{y} = 4 \) with respect to \(x\). Remember that \(4\) is a constant, so its derivative is \(0\).

Step 2 - Apply the chain rule

Differentiate \(\sqrt{x}\) and \(\sqrt{y}\) using the chain rule: \ \ For \(\sqrt{x}\): \ \ \(\frac{d}{dx}(x^{1/2}) = \frac{1}{2\sqrt{x}} \). \ \ For \(\sqrt{y}\), since \(y\) is a function of \(x\): \ \ \(\frac{d}{dx}(y^{1/2}) = \frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx} \).

Step 3 - Combine the derivatives

Combine the derivatives from Step 2: \ \ \( \frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx} = 0 \).

Step 4 - Solve for \(\frac{dy}{dx}\)

Solve the equation from Step 3 for \(\frac{dy}{dx}\): \ \ Subtract \( \frac{1}{2\sqrt{x}} \) from both sides: \ \ \( \frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx} = -\frac{1}{2\sqrt{x}} \). \ \ Multiply both sides by \(2\sqrt{y}\): \ \ \( \frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}} \).

Key Concepts

Chain Rule

The chain rule is a fundamental tool in calculus used to differentiate composite functions. Imagine function compositions like nested Russian dolls; to differentiate them, you need to carefully peel away each layer.
Consider the function \(\frac{d}{dx}(y^{1/2})\). Because \(y\) is a function of \(x\), we can't differentiate it directly. Instead, we use the chain rule: \[ \frac{d}{dx}(y^{1/2}) = \frac{1}{2\sqrt{y}} \frac{dy}{dx} \] Here, \(\frac{1}{2\sqrt{y}}\) is the derivative of \(y^{1/2}\) treated as an external variable times \(\frac{dy}{dx}\), accounting for \(y\)'s internal dependence on \(x\). By mastering the chain rule, you unlock the ability to differentiate a vast array of complex functions.

Differentiation

Differentiation is the process of finding a derivative, which represents the rate of change of a function. In this context, we need to differentiate both sides of the equation \(\sqrt{x} + \sqrt{y} = 4\).
Differentiation allows us to see how the variables \(x\) and \(y\) change relative to each other. For instance, the derivative of \(\sqrt{x}\) with respect to \(x\) is \(\frac{1}{2\sqrt{x}} \). Because \(4\) is a constant, its derivative is zero.
During implicit differentiation, we differentiate every term in the given equation, recognizing that \(y\) is a function of \(x\). This step helps to systematically track how changes in \(x\) affect \(y\).

Solve for Derivative

To solve for the derivative, we isolate \(\frac{dy}{dx}\). After differentiating the equation \(\sqrt{x}+\sqrt{y}=4\) and applying the chain rule, we get: \[ \frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \frac{dy}{dx} = 0 \]
We then isolate \(\frac{dy}{dx}\) by subtracting the \(\frac{1}{2\sqrt{x}}\) term from both sides, resulting in: \[ \frac{1}{2\sqrt{y}} \frac{dy}{dx} = -\frac{1}{2\sqrt{x}} \]
Finally, by multiplying both sides by \(2\sqrt{y}\) to clear the fraction, we find: \[ \frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}} \]
This represents the rate at which \(y\) changes with respect to \(x\).

Functions of Multiple Variables

In calculus, functions of multiple variables involve more than one independent variable. For instance, \(y\) could depend on \(x\) and another variable. In the equation \(\sqrt{x} + \sqrt{y} = 4\), \(x\) and \(y\) are interdependent.
When differentiating such functions, each variable's partial derivative is considered. Implicit differentiation allows you to handle these interdependent relationships by treating one variable as an implicit function of another.
By understanding how to navigate and differentiate these relationships, you gain the power to analyze more complex systems in mathematics and the real world.