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Chapter 2: Problem 13

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

Short Answer

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

Step by step solution

01

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: \[\frac{d}{dx}\left(\sqrt{x}\right) + \frac{d}{dx}\left(\sqrt{y}\right) = \frac{d}{dx}(1)\]
02

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: \[\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}}\frac{dy}{dx} = 0\]
03

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 \(2\sqrt{y}\): \[\frac{dy}{dx} = -\frac{1}{2\sqrt{x}} \cdot 2\sqrt{y}\]
04

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: \[\frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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鈥攕ince \( y \) is a function of \( x \)鈥攖he 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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