Chapter 3: Problem 13
Differentiate implicily to find \(d y / d x\). $$ x^{2}-y^{2}=16 $$
Short Answer
Expert verified
The derivative \( \frac{dy}{dx} \) is \( \frac{x}{y} \).
Step by step solution
01
Understand the Exercise
The exercise asks to differentiate the equation implicitly in order to find \( \frac{dy}{dx} \). The given equation is \( x^2 - y^2 = 16 \).
02
Differentiate Both Sides
Differentiate both sides of the equation with respect to \( x \). Applying the differentiation rule, we get:\[ \frac{d}{dx} (x^2 - y^2) = \frac{d}{dx} (16) \]
03
Apply the Chain Rule
Differentiate the left side. For \( x^2 \), the derivative is straightforward, yielding \( 2x \). For \( -y^2 \), use the chain rule: \[ -\frac{d}{dx} (y^2) = -2y \frac{dy}{dx} \].Therefore, you have:\[ 2x - 2y \frac{dy}{dx} = 0 \]
04
Solve for \( \frac{dy}{dx} \)
Isolate \( \frac{dy}{dx} \) on one side of the equation. First, add \( 2y \frac{dy}{dx} \) to both sides:\[ 2x = 2y \frac{dy}{dx} \]Next, divide both sides by \( 2y \):\[ \frac{2x}{2y} = \frac{dy}{dx} \].Simplify the fraction:\[ \frac{x}{y} = \frac{dy}{dx} \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Implicit Differentiation
Implicit differentiation is used when you have an equation involving both x and y, and you need to find the derivative, \frac{dy}{dx}\. In some cases, y is not isolated on one side of the equation, hence the need for implicit differentiation. Here's how it's done:
1. Differentiate both sides of the equation with respect to x. Keep in mind that y is also a function of x.
2. Apply standard differentiation rules like the chain rule to differentiate terms involving y.
In the given problem, the equation is \( x^2 - y^2 = 16 \). When differentiating both sides with respect to x, treat y as a function of x. This requires employing the chain rule for differentiating \( y^2 \).
In mathematics, many functions are not given explicitly as \( y = f(x) \). Instead, they may be given implicitly as some relation between x and y. Implicit differentiation allows us to find the derivative in such cases.
Finally, isolating \( \frac{dy}{dx} \) by algebraic manipulation gives us the result.
1. Differentiate both sides of the equation with respect to x. Keep in mind that y is also a function of x.
2. Apply standard differentiation rules like the chain rule to differentiate terms involving y.
In the given problem, the equation is \( x^2 - y^2 = 16 \). When differentiating both sides with respect to x, treat y as a function of x. This requires employing the chain rule for differentiating \( y^2 \).
In mathematics, many functions are not given explicitly as \( y = f(x) \). Instead, they may be given implicitly as some relation between x and y. Implicit differentiation allows us to find the derivative in such cases.
Finally, isolating \( \frac{dy}{dx} \) by algebraic manipulation gives us the result.
Chain Rule
The chain rule is a fundamental concept in calculus used to differentiate compositions of functions. It states that if a variable y depends on a variable u, which in turn depends on x, then y also depends on x, and the derivative of y with respect to x can be found using:
\[ \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \].
In the equation from the exercise, for the term \( y^2 \), y is a function of x. By the chain rule, the derivative of \( y^2 \) with respect to x is:
\[ \frac{d}{dx} (y^2) = 2y \frac{dy}{dx} \].
This step is crucial because it allows us to differentiate terms where y is implicitly a function of x. The chain rule was applied in this problem to correctly handle the differentiation of \( y^2 \) and obtain the relationship between x and y's derivatives.
\[ \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \].
In the equation from the exercise, for the term \( y^2 \), y is a function of x. By the chain rule, the derivative of \( y^2 \) with respect to x is:
\[ \frac{d}{dx} (y^2) = 2y \frac{dy}{dx} \].
This step is crucial because it allows us to differentiate terms where y is implicitly a function of x. The chain rule was applied in this problem to correctly handle the differentiation of \( y^2 \) and obtain the relationship between x and y's derivatives.
Differentiation and Derivatives
Differentiation is the process of finding the derivative of a function. The derivative, denoted as \( \frac{dy}{dx} \), measures the rate at which a function is changing at any point. It gives us the slope of the tangent line to the curve of the function at any given point.
Key points about differentiation:
1. To differentiate \( x^2 \) w.r.t x, we use the power rule: \( \frac{d}{dx} (x^2) = 2x \).
2. To differentiate \( -y^2 \), we need the chain rule, which we discussed earlier. The term differentiates to \( -2y \frac{dy}{dx} \), incorporating both the chain rule and the power rule.
By combining these rules, we can differentiate complex equations and relations. The final simplification \( x = y \frac{dy}{dx} \) isolates \( \frac{dy}{dx} \), giving us the derivative of y with respect to x.
Key points about differentiation:
- The power rule: If \( f(x) = x^n \), then \( f'(x) = nx^{n-1} \).
- The sum rule: The derivative of a sum of functions is the sum of their derivatives.
- The difference rule: The derivative of a difference of functions is the difference of their derivatives.
1. To differentiate \( x^2 \) w.r.t x, we use the power rule: \( \frac{d}{dx} (x^2) = 2x \).
2. To differentiate \( -y^2 \), we need the chain rule, which we discussed earlier. The term differentiates to \( -2y \frac{dy}{dx} \), incorporating both the chain rule and the power rule.
By combining these rules, we can differentiate complex equations and relations. The final simplification \( x = y \frac{dy}{dx} \) isolates \( \frac{dy}{dx} \), giving us the derivative of y with respect to x.