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Chapter 3: Problem 19

Differentiate implicily to find \(d y / d x\). $$ \sqrt{x}+\sqrt{y}=1 $$

Short Answer

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

Step by step solution

01

Differentiate Both Sides with Respect to x

Differentiate each term on both sides separately. Remember that y is a function of x.
02

Differentiate the First Term

The first term is \( \sqrt{x} \). Applying the chain rule, the derivative of \( \sqrt{x} \) with respect to \( x \) is \( \frac{1}{2\sqrt{x}} \).
03

Differentiate the Second Term

The second term is \( \sqrt{y} \). Using the chain rule, its derivative with respect to \( x \) is \( \frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx} \).
04

Differentiate the Right Side

The right side of the equation is simply 1. The derivative of a constant is 0.
05

Combine the Results

Combine the derivatives obtained: \( \frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx} = 0 \).
06

Solve for \( \frac{dy}{dx} \)

Isolate \( \frac{dy}{dx} \) by solving the resulting equation: \[ \frac{dy}{dx} = - \frac{\sqrt{y}}{\sqrt{x}} \]

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Key Concepts

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

Chain Rule in Implicit Differentiation
When differentiating implicitly, the chain rule is a crucial tool. This is because we recognize that some variables, like y in our example, depend on another variable, x. For instance, when differentiating \( \sqrt{y} \ \) with respect to x, we first treat \( y \ \) as a function of x. We then apply the chain rule:
  • Differentiate the outer function: \( \sqrt{y} \ \), which gives \( \frac{1}{2 \sqrt{y}} \ \)
  • Differentiate the inner function (y) with respect to x: \( \frac{dy}{dx} \ \)
  • Combine both results: \( \frac{1}{2 \sqrt{y}} \cdot \frac{dy}{dx} \ \)
This process simplifies complex derivatives into manageable parts.
Understanding Derivatives
Derivatives measure how a function changes as its input changes. In implicit differentiation, you need to find the rate at which y changes with respect to x. In our problem \( \sqrt{x} + \sqrt{y} = 1 \ \), differentiating each term separately ensures precision.
  • For \( \sqrt{x} \ \), the standard derivative is \( \frac{1}{2 \sqrt{x}} \ \)
  • For \( \sqrt{y} \ \), we again use the chain rule to get \( \frac{1}{2 \sqrt{y}} \cdot \frac{dy}{dx} \ \)
  • Constants like 1 differentiate to 0
Mastering these steps builds a solid foundation for solving more complex problems.
Functions of x and their Relationships
Functions describe the relationship between variables. In our exercise, \( x \ \) and \( y \ \) are related by \( \sqrt{x} + \sqrt{y} = 1 \ \). This means any change in \( x \ \) affects \( y \ \). Recognizing this relationship is key in implicit differentiation.
  • When x changes, we need to adjust y, maintaining the equation balance
  • This relationship guides us in applying the chain rule correctly
  • Ensuring we differentiate each term with respect to x
Understanding these connections helps you visualize how variables interact within an equation.

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