Question

Show that \(\frac{d y}{d x}\) is the same for each of the following implicitly defined functions. (a) \(x y=1\) (b) \(x^{2} y^{2}=1\) (c) \(\sin (x y)=1\) (d) \(\ln (x y)=1\)

Short answer

For all cases, \(\frac{dy}{dx} = -\frac{y}{x}\).

Step-by-step solution

Differentiate the Implicit Function (a)

For the function \(xy = 1\), use implicit differentiation. Differentiating both sides with respect to \(x\) gives:\[ \frac{d}{dx}(xy) = \frac{d}{dx}(1) \]Using the product rule on the left side, we get:\[ y + x\frac{dy}{dx} = 0 \]Isolate \(\frac{dy}{dx}\):\[ \frac{dy}{dx} = -\frac{y}{x} \]

Differentiate the Implicit Function (b)

For the function \(x^2 y^2 = 1\), apply implicit differentiation. Differentiate both sides with respect to \(x\):\[ \frac{d}{dx}(x^2 y^2) = \frac{d}{dx}(1) \]Applying the product rule and chain rule, we have:\[ 2xy^2 + 2x^2y \frac{dy}{dx} = 0 \]Rearrange to solve for \(\frac{dy}{dx}\):\[ 2x^2 y \frac{dy}{dx} = -2x y^2 \]\[ \frac{dy}{dx} = -\frac{y}{x} \]

Differentiate the Implicit Function (c)

For \(\sin(xy) = 1\), differentiate both sides with respect to \(x\):\[ \frac{d}{dx}(\sin(xy)) = \frac{d}{dx}(1) \]The derivative of sine is cosine, so applying the chain rule:\[ \cos(xy)\cdot(x\frac{dy}{dx} + y) = 0 \]Since \(\cos(xy) eq 0\) (as \(\sin(xy) = 1\) only if \((xy) = \frac{\pi}{2} + n\pi\) where \(n\) is an integer), solve for \(\frac{dy}{dx}\):\[ x\frac{dy}{dx} = -y \]\[ \frac{dy}{dx} = -\frac{y}{x} \]

Differentiate the Implicit Function (d)

For \(\ln(xy) = 1\), differentiate both sides with respect to \(x\):\[ \frac{d}{dx}(\ln(xy)) = \frac{d}{dx}(1) \]The derivative of \(\ln(u)\) is \(\frac{1}{u}\cdot\frac{du}{dx}\), so apply the chain rule:\[ \frac{1}{xy} \cdot (x\frac{dy}{dx} + y) = 0 \]Rearrange and simplify to find \(\frac{dy}{dx}\):\[ x\frac{dy}{dx} + y = 0 \]\[ x\frac{dy}{dx} = -y \]\[ \frac{dy}{dx} = -\frac{y}{x} \]

Key concepts

Product Rule

When you come across expressions involving products of two functions, the product rule is a powerful tool in differentiation. The essence of the product rule is that when differentiating a product of two functions—let's call them \(u\) and \(v\)—the derivative of the product \(uv\) is not merely the product of their derivatives, but rather:

• The derivative of the first function \(u\) multiplied by the second function \(v\), plus
• The first function \(u\) multiplied by the derivative of the second function \(v\).

In formulaic terms, the product rule is expressed as:\[ \frac{d}{dx}[u v] = u'v + uv' \]This rule ensures that each component of the product is accounted for and the way in which one function changes affects the product overall. In the original exercise, for equations like \(xy = 1\), the product \(xy\) involves the multiplication of two variables. By utilizing the product rule, we distinctly separate how each variable affects the differentiation process.

Chain Rule

The chain rule is indispensable when dealing with compositions of functions. It allows us to differentiate a function that is made up by nesting one function inside another. Say you have a composite function written as \(f(g(x))\). To differentiate this, you adopt the chain rule formula:

• First, differentiate the outer function \(f\) as if the inner function \(g(x)\) were just a variable.
• Then, multiply the result by the derivative of the inner function \(g\).

Mathematically, it looks like:\[ \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \]In the original exercise, functions like \( \sin(xy) = 1 \) and \( \ln(xy) = 1 \) are implicitly defined, where the chain rule is used to handle the composite nature of the functions \( \sin \) and \( \ln \), which have \(xy\) as their inner functions.

Differentiation

Differentiation is the process of finding the rate at which a function is changing at any given point, typically expressed as the derivative of the function. In essence, it responds to the question: "How does one quantity change as another quantity changes?"
In calculus, differentiation is used to find the slope of a function at any point on its curve. It's like asking, "If this point belonged to a line, what would be the angle of its slope?"
During implicit differentiation, as seen in the exercises such as \(x^2y^2 = 1\) and \( \ln(xy) = 1 \), you differentiate all terms with respect to \(x\), yet not all of these terms are neatly isolated. Hence, implicit differentiation involves getting the derivative of \(y\) as a function of \(x\), managing the two variables' interaction throughout.

Derivatives

A derivative is a fundamental concept in calculus describing the rate of change of a function. It's akin to finding the velocity of a car at a specific moment: it tells us how fast something is moving or changing at an instant.

• The notation \( \frac{dy}{dx} \) is used to represent the derivative of \(y\) with respect to \(x\), symbolizing this rate of change.
• In the implicit functions given, determining \( \frac{dy}{dx} \) means unraveling how \(y\) changes as \(x\) changes.

In the exercises provided, all results boil down to \( \frac{dy}{dx} = -\frac{y}{x} \), a clean expression showing the balance of change between \(x\) and \(y\) thanks to the use of implicit differentiation, the product rule, and the chain rule.