Problem 13

Question

Find dy/dx by implicit differentiation. $$ \sqrt{x}+\sqrt{y}=1 $$

Step-by-Step Solution

Verified

Answer

The short answer is: \[\frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}}.\]

1Step 1: Differentiate both sides of the equation with respect to x

First, we need to differentiate both sides of the given equation, $\sqrt{x} + \sqrt{y} = 1$, concerning x using the chain rule. Remember that when differentiating y with respect to x, we have to multiply by dy/dx. Differentiating both sides of the equation, we have: \[\frac{d}{dx}\left(\sqrt{x}\right) + \frac{d}{dx}\left(\sqrt{y}\right) = \frac{d}{dx}(1)\]

2Step 2: Apply the chain rule and differentiation rules

On the left side, we use the chain rule and the fact that the derivative of $\sqrt{x}$ is $\frac{1}{2\sqrt{x}}$. To differentiate $\sqrt{y}$ concerning x, we use the chain rule again and the fact that the derivative of $\sqrt{y}$ concerning y is $\frac{1}{2\sqrt{y}}$. Finally, the right side is a constant, so its derivative concerning x is 0. Applying these rules, we have: \[\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}}\frac{dy}{dx} = 0\]

3Step 3: Solve for dy/dx

To find the value of dy/dx, we need to isolate dy/dx term in the equation. Let's first subtract $\frac{1}{2\sqrt{x}}$ from both sides of the equation: \[\frac{1}{2\sqrt{y}}\frac{dy}{dx} = -\frac{1}{2\sqrt{x}}\] Now, to isolate dy/dx, we need to multiply both sides of the equation by the reciprocal of $\frac{1}{2\sqrt{y}}$, which is $2\sqrt{y}$: \[\frac{dy}{dx} = -\frac{1}{2\sqrt{x}} \cdot 2\sqrt{y}\]

4Step 4: Simplify the result

Simplifying the expression gives us: \[\frac{dy}{dx} = -\sqrt{y}\div\sqrt{x}\] So, the derivative of y with respect to x or dy/dx is given by: \[\frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}}\]

Key Concepts

Chain RuleDerivativeCalculusEquation Solving

Chain Rule

The chain rule is a fundamental principle in calculus. It allows us to differentiate compositions of functions. Simply put, it helps us find the derivative of a function that is inside another function. In our exercise, this is particularly useful for differentiating terms like $ \sqrt{y} $ with respect to $ x $.

For the function $ \sqrt{y} $, first find the derivative with respect to $ y $, which is $ \frac{1}{2\sqrt{y}} $.
Next, multiply it by the derivative of $ y $ with respect to $ x $, which is $ \frac{dy}{dx} $.

This yields a term $ \frac{1}{2\sqrt{y}} \frac{dy}{dx} $, showing how the change in $ y $ influences its effect when compared to $ x $.

Derivative

A derivative represents the rate at which one quantity changes with respect to another. In simpler terms, it gives us the slope of a function at any given point. In our problem, we are finding $ \frac{dy}{dx} $, the derivative of $ y $ with respect to $ x $.

The derivative of $ \sqrt{x} $ with respect to $ x $ is $ \frac{1}{2\sqrt{x}} $, highlighting the rate at which $ \sqrt{x} $ changes as $ x $ changes.
Similarly, using implicit differentiation—since $ y $ is a function of $ x $—the derivative of $ \sqrt{y} $ includes $ \frac{dy}{dx} $, indicating the relationship between changes in $ x $ and $ y $.

In the equation $ \frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \frac{dy}{dx} = 0 $, solving for $ \frac{dy}{dx} $ helps us understand how both variables relate.

Calculus

Calculus is a branch of mathematics focused on change and motion. It divided into two main parts: differentiation and integration. In this exercise, we utilize differentiation, particularly implicit differentiation, to solve an equation involving $ x $ and $ y $.

Implicit differentiation allows us to find derivatives even when $ y $ is not isolated explicitly as a function of $ x $.
Using this method, we differentiate both sides of the equation $ \sqrt{x} + \sqrt{y} = 1 $ concerning $ x $.
Calculus principles guide us in understanding how variables like $ y $ change when not given explicitly.

Implicit differentiation helps capture the dynamic relationships present in equations involving more than one variable, as seen by the relationship between $ x $ and $ y $ in our problem.

Equation Solving

Solving equations often requires rearranging expressions to isolate a specific variable. In this problem, we solve for $ \frac{dy}{dx} $, the derivative of $ y $ with respect to $ x $.

Starting from $ \frac{1}{2\sqrt{y}} \frac{dy}{dx} = -\frac{1}{2\sqrt{x}} $, notice the goal is to isolate $ \frac{dy}{dx} $.
We multiply both sides by $ 2\sqrt{y} $ to rid the fraction, which gives $ \frac{dy}{dx} = -\sqrt{y} \div \sqrt{x} $.
Simplification leads to a cleaner, computable expression showing how changes in $ y $ are proportional to changes in $ x $.

Understanding the process of algebraically manipulating expressions ensures clarity when determining relationships in calculus problems.

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