Question

Assuming that each equation defines a differentiable function of \(x\), find \(D_{x} y\) by implicit differentiation. \(y^{2}-x^{2}=1\)

Short answer

\(\frac{dy}{dx} = \frac{x}{y}\)

Step-by-step solution

Differentiate both sides with respect to x

The given equation is \(y^2 - x^2 = 1\). We need to find the derivative \( \frac{dy}{dx} \) using implicit differentiation. First, differentiate both sides of the equation with respect to \(x\). This gives: \[ \frac{d}{dx}(y^2) - \frac{d}{dx}(x^2) = \frac{d}{dx}(1). \]

Apply the chain rule

Differentiate \(y^2\) using the chain rule: \( (2y) \cdot \frac{dy}{dx} \), and differentiate \(x^2\) normally: \(2x\). The right side, which is a constant, becomes 0. So we have: \[ 2y \cdot \frac{dy}{dx} - 2x = 0. \]

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

Rearrange the differentiated equation to solve for \(\frac{dy}{dx}\): \[ 2y \cdot \frac{dy}{dx} = 2x \]Divide both sides by \(2y\) (assuming \(y eq 0\)) to isolate \(\frac{dy}{dx}\): \[ \frac{dy}{dx} = \frac{x}{y}. \]

Key concepts

chain rule

Implicit differentiation often requires the use of the chain rule, a fundamental concept in calculus. The chain rule is essential when differentiating composite functions—functions within functions.

Consider a scenario where we have a function, say \( y^2 \). To understand why the chain rule is necessary here, remember that \( y \) itself is a function of \( x \). This transforms \( y^2 \) into a composition: a function \( y(x) \) inside another, which is squaring the result. Therefore, when differentiating \( y^2 \), we can't treat \( y \) as a constant; instead, we apply the chain rule.

In this particular exercise, using the chain rule gives us \( 2y \cdot \frac{dy}{dx} \). Why? Because when differentiating a function like \( f(g(x)) \), the rule states we must multiply the derivative of the outer function—\( 2y \), as it was derived from \( y^2 \) —by the derivative of the inner function, \( \frac{dy}{dx} \), resulting from how \( y \) changes with \( x \). This multiplication accounts for any variation in \( y \) as \( x \) shifts, reflecting real-world dependencies in calculated derivatives.

differentiable function

Differentiable functions are pivotal to calculus, describing continuous and smoothly varying changes. If a function is differentiable, it means that at every point in its domain, there's a unique tangent line, highlighting its predictability and continuity.

In the problem at hand, the assumption that each equation defines a differentiable function of \( x \) is crucial. What does this mean practically? It ensures that as we compute derivatives—like the example of \( y = f(x) \)—the functions involved are nicely behaved without abrupt jumps or discontinuities.

These properties grant us the ability to apply differentiation rules, like the chain rule, with confidence. As we have considered \( y^2 - x^2 = 1 \), we imagine a curve representing a smooth transition. Each step in differentiation depends on this core assumption, providing a reliable framework to solve for \( \frac{dy}{dx} \). Without differentiability, the function could be unpredictable, making standard calculus methods, such as implicit differentiation, ineffective.

derivative calculation

Derivative calculation is the cornerstone of calculus, used to understand how functions change. When confronted with an equation involving several variables, like \( y^2 - x^2 = 1 \), derivatives help us grasp how one variable varies in response to another.

The process begins by taking derivatives with respect to a chosen variable, typically \( x \). For implicit differentiation, both sides of the equation are differentiated simultaneously, maintaining the balance of the equation. On differentiating both sides of \( y^2 - x^2 = 1 \), you reach:

• \( \frac{d}{dx}(y^2) \) which uses the chain rule: \( (2y) \cdot \frac{dy}{dx} \)
• \( \frac{d}{dx}(x^2) = 2x \)
• \( \frac{d}{dx}(1) = 0 \) since constants have no derivative

Altogether, these calculated derivatives form an equation: \( 2y \cdot \frac{dy}{dx} - 2x = 0 \). Solving this involves algebra to isolate \( \frac{dy}{dx} \), leaving us with \( \frac{dy}{dx} = \frac{x}{y} \).

This final expression, often expressed as the slope of the curve at any \( (x, y) \) point, reveals the relationship between \( x \) and \( y \) changes. This results from the elegant dance of calculation rules and function differentiability, orchestrated by mathematics to explore and describe dynamics thoroughly.