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Magazine Chapter 3: Problem 1
Find \(d y / d x .\) Assume \(a, b, c\) are constants. $$x^{2}+y^{2}=\sqrt{7}$$
Short Answer
Expert verified
\(\frac{dy}{dx} = -\frac{x}{y}\)
Step by step solution
01
Write the Given Equation
The equation given in the exercise is \(x^2 + y^2 = \sqrt{7}\).
02
Differentiate Both Sides
Differentiate both sides of the equation with respect to \(x\). For the left side, use the power rule and implicit differentiation. We get: \(\frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}(\sqrt{7})\).
03
Apply Derivative to Each Term
Applying the derivative on the left side yields:\(\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = 0\). For \(x^2\), the derivative is \(2x\). For \(y^2\), use implicit differentiation to get \(2y \frac{dy}{dx}\).
04
Combine Derivatives
The equation from Step 3 is now:\[2x + 2y \frac{dy}{dx} = 0\].
05
Solve for \(\frac{dy}{dx}\)
Rearrange the equation to isolate \(\frac{dy}{dx}\):\[2y \frac{dy}{dx} = -2x\]. Divide both sides by \(2y\): \[\frac{dy}{dx} = -\frac{x}{y}\].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding the Derivative
The derivative is a fundamental concept in calculus, representing the rate at which a function is changing at any given point. It is often considered the slope of the tangent line to the curve of a function at a point. In our exercise, the derivative helps us understand how the variables \( x \) and \( y \) change in relation to each other given the constraint \( x^2 + y^2 = \sqrt{7} \).
- Essentially, when we differentiate, we are looking at the "instantaneous rate of change."
- The derivative of a function is denoted as \( \frac{dy}{dx} \), which reads as "the derivative of \( y \) with respect to \( x \)."
Applying the Power Rule
The power rule is one of the simplest derivative rules and is used to differentiate power functions of the form \( x^n \). According to the power rule, if you have \( x^n \), the derivative is \( nx^{n-1} \). This makes it straightforward to handle terms like \( x^2 \) or \( y^2 \) in our equation.
- For \( x^2 \), applying the power rule gives the derivative \( 2x \).
- For \( y^2 \), it is essential to remember that \( y \) is a function of \( x \), which involves implicit differentiation.
Diving into Implicit Differentiation
Implicit differentiation is a technique used when a relationship between variables is given implicitly rather than explicitly (i.e., one variable is not isolated on one side of the equation).
For the equation \( x^2 + y^2 = \sqrt{7} \):
For the equation \( x^2 + y^2 = \sqrt{7} \):
- We differentiate both sides with respect to \( x \). The derivative of \( y^2 \) requires implicit differentiation because \( y \) is considered a function of \( x \).
- The process involves taking the derivative of each term with respect to \( x \), which for \( y^2 \) leads to \( 2y \frac{dy}{dx} \).
Grasping Differentiation Concepts
Differentiation is the overall process of finding a derivative. It is a key operation in calculus that is used to understand how functions change over intervals. Differentiation can reveal critical points, rates of change, and trends within mathematical relationships.
- The process requires an understanding of rules like the power rule, as well as techniques such as implicit differentiation when functions are not straightforward.
- Differentiation is integral to solving real-world problems that involve continuous change, such as physics and engineering contexts.
- Beyond the basic principles, practice with various types of differentiation will enhance problem-solving skills.
