Problem 7 - Single Variable Calculus

Question

Find dy/dx by implicit differentiation. } $$ \frac{1}{x}+\frac{1}{y}=1 $$

Step-by-Step Solution

Answer

The short version of the answer for the given implicit differentiation problem is: \(\frac{dy}{dx} = \frac{y^2}{x^2}\).

Step 1: Differentiate each term with respect to x

Recall that when differentiating with respect to x, we'll treat y as a function of x, meaning we'll need to use the chain rule: \(\frac{d}{dx}(f(g(x))) = f'(g(x)) \cdot g'(x)\). Starting with \(\frac{1}{x}\): \(\frac{d}{dx}\left(\frac{1}{x}\right) = -\frac{1}{x^2}\) Now, differentiate \(\frac{1}{y}\) with respect to x: \(\frac{d}{dx}\left(\frac{1}{y}\right) = -\frac{1}{y^2} \cdot \frac{dy}{dx}\) Finally, differentiate the constant \(1\) with respect to x: \(\frac{d}{dx}(1) = 0\)

Step 2: Rewrite the equation with the derivatives

Substitute the derivatives we found in Step 1 back into the original equation: \(-\frac{1}{x^2} - \frac{1}{y^2}\frac{dy}{dx}=0\)

Step 3: Solve for dy/dx

Now, we need to solve for \(\frac{dy}{dx}\). First, isolate the \(\frac{dy}{dx}\) term: \(\frac{1}{y^2} \frac{dy}{dx} = \frac{1}{x^2}\) Finally, multiply by \(y^2\) to solve for \(\frac{dy}{dx}\): \(\frac{dy}{dx} = \frac{y^2}{x^2}\) So, \(\frac{dy}{dx} = \frac{y^2}{x^2}\).

Key Concepts

Chain Rule

The chain rule is an essential concept in calculus, particularly when dealing with composite functions. This rule helps us find the derivative of functions that are nested inside one another.
In our exercise, we encounter this situation, especially with the term \( \frac{1}{y} \). Here, \(y\) is considered a function of \(x\), even though it’s not explicitly stated.

• To apply the chain rule properly, you think of \( \frac{1}{y} \) as a composition of the function \( f(y) = \frac{1}{y} \) and the implicit function \( y(x) \).
• The chain rule tells us that the derivative of \( f(y(x)) \) with respect to \(x\) is \( f'(y) \cdot y'(x) \). This becomes: \(-\frac{1}{y^2} \cdot \frac{dy}{dx}\).

This technique allows us to differentiate terms with respect to \(x\), even when \(y\) is present.

Derivative

A derivative represents the rate at which a function is changing at any given point. It's like finding the 'slope' of a function at a particular point.
For functions involving variables in the denominator, like \( \frac{1}{x} \), the power rule becomes useful. Remember that \( \frac{1}{x} \) can be rewritten as \(x^{-1}\).

• Using the power rule, the derivative \( \frac{d}{dx}\left( x^{-1} \right) \) results in \(-x^{-2}\).
• Applying this understanding helps us compute each term's derivative in the context of our original equation.

Solve for dy/dx

After computing the derivatives, the next task is to "solve for \(\frac{dy}{dx}\)." This means isolating the \(\frac{dy}{dx}\) term on one side of the equation.

• Start by rewriting the differentiated equation, to simplify the algebraic steps. For example, bring terms involving \(\frac{dy}{dx}\) to one side. In our case: \( \frac{1}{y^2} \frac{dy}{dx} = \frac{1}{x^2} \).
• Then, to isolate \(\frac{dy}{dx}\), multiply both sides by \(y^2\), giving you the final result: \(\frac{dy}{dx} = \frac{y^2}{x^2}\).

It's important to follow these algebraic steps carefully to ensure you correctly solve for \(\frac{dy}{dx}\).

Differentiation

Differentiation is the mathematical process of finding a derivative. It is the core operation in calculus used to understand how things change. In our exercise, we specifically used a technique called "implicit differentiation."

• Implicit differentiation is useful when you have equations that are not solved for either variable. For example, our given equation is \( \frac{1}{x} + \frac{1}{y} = 1 \), where \(y\) is not isolated.
• When performing implicit differentiation, always remember to differentiate each term with respect to \(x\). Terms involving \(y\) require the chain rule since \(y\) is treated as a function of \(x\).

By using implicit differentiation, we can find \(\frac{dy}{dx}\), thus understanding how \(y\) changes relative to \(x\). This reveals the deeper interrelationship between the two variables as prescribed by their equation.