Problem 32 - Thomas Calculus

Question

Use implicit differentiation to find \(d y / d x\) in Exercises \(19-32\) $$ y^{2} \cos \left(\frac{1}{y}\right)=2 x+2 y $$

Step-by-Step Solution

Answer

\(\frac{dy}{dx} = \frac{2}{2y \cos \left(\frac{1}{y}\right) - \sin \left(\frac{1}{y}\right)}\).

Step 1: Differentiate Both Sides with Respect to x

Start by differentiating the left-hand side of the equation. Use the product rule on the expression \(y^{2} \cos \left(\frac{1}{y}\right)\). The expression is a product of \(y^2\) and \(\cos \left(\frac{1}{y}\right)\). The product rule states \((uv)' = u'v + uv'\). Here, \(u = y^2\) and \(v = \cos \left( \frac{1}{y} \right)\). Differentiate \(u\) and \(v\) with respect to \(x\).

Step 2: Apply the Chain Rule

While differentiating \(u = y^2\), apply the chain rule: \(\frac{d}{dx}(y^2) = 2y \cdot \frac{dy}{dx}\). For \(v = \cos \left(\frac{1}{y}\right)\), use the chain rule on \(\frac{d}{dy}\): \(\frac{d}{dx}\cos \left(\frac{1}{y}\right) = -\sin \left(\frac{1}{y}\right) \cdot \left(-\frac{1}{y^2}\right) \cdot \frac{dy}{dx}\).

Step 3: Combine the Derivatives Using Product Rule

Combine the results from the chain rule application: \(\frac{d}{dx} [y^{2} \cos \left(\frac{1}{y}\right)] = 2y \cos \left(\frac{1}{y}\right) \frac{dy}{dx} - y^{-2} \sin \left(\frac{1}{y}\right) \cdot y^2 \frac{dy}{dx}\).

Step 4: Differentiate the Right-Hand Side

Differentiate the right-hand side, \(2x + 2y\), with respect to \(x\): \(\frac{d}{dx}(2x + 2y) = 2 + 2 \frac{dy}{dx}\).

Step 5: Formulate the Full Equation

Combine the derivatives from both sides to form one equation: \[2y \cos \left(\frac{1}{y}\right) \frac{dy}{dx} - y^{-2} \sin \left(\frac{1}{y}\right) \cdot y^2 \frac{dy}{dx} = 2 + 2 \frac{dy}{dx}.\]

Step 6: Solve for \(\frac{dy}{dx}\)

Collect all terms involving \(\frac{dy}{dx}\) on one side and factor them out:\[\left(2y \cos \left(\frac{1}{y}\right) - \sin \left(\frac{1}{y}\right)\right) \frac{dy}{dx} = 2.\] Divide both sides by \(2y \cos \left(\frac{1}{y}\right) - \sin \left(\frac{1}{y}\right)\) to solve for \(\frac{dy}{dx}\):\[\frac{dy}{dx} = \frac{2}{2y \cos \left(\frac{1}{y}\right) - \sin \left(\frac{1}{y}\right)}.\]

Key Concepts

Chain Rule: Understanding and Application

The chain rule is a fundamental tool in calculus used to differentiate composite functions. It allows us to find the derivative of a function that can be broken down into two or more nested functions. Let’s consider a composite function \( f(g(x)) \), where \( f \) is an outer function and \( g \) is an inner function. The chain rule states that the derivative of this composite function is:

• \( \frac{d}{dx} f(g(x)) = f'(g(x)) \times g'(x) \)

To illustrate this, look at the solution where we needed the chain rule for the function \( y^2 \) and \( \cos(\frac{1}{y}) \). When differentiating \( y^2 \) with respect to \( x \), we applied the chain rule since \( y \) is a function of \( x \). The calculation was structured as:

• \( \frac{d}{dx}(y^2) = 2y \cdot \frac{dy}{dx} \)

This method was also applied to find the derivative of \( \cos(\frac{1}{y}) \). Here, the inner function is \( \frac{1}{y} \), and the chain rule applied gives:

• \( -\sin(\frac{1}{y}) \cdot \left(-\frac{1}{y^2}\right) \cdot \frac{dy}{dx} \)

Using the chain rule helps in breaking down the complexities of differentiating these composite structures, simplifying our calculation process.

Product Rule: Differentiating Products of Functions

The product rule is another vital rule in calculus, specifically for differentiating products of two functions. This rule states that if you have a function \( h(x) = u(x) \times v(x) \), the derivative is given by:

• \( h'(x) = u'(x)v(x) + u(x)v'(x) \)

In this solution, we encounter a product: \( y^2 \cdot \cos\left(\frac{1}{y}\right) \). Here:

• \( u = y^2 \)
• \( v = \cos\left(\frac{1}{y}\right) \)

The product rule was used as:

• \( \frac{d}{dx}[y^2 \cdot \cos(\frac{1}{y})] = (2y \cdot \frac{dy}{dx}) \cdot \cos(\frac{1}{y}) + y^2 \cdot \left(-\sin(\frac{1}{y}) \cdot \left(-\frac{1}{y^2}\right) \cdot \frac{dy}{dx}\right) \)

Notice how each term is differentiated as per its necessity, with the chain rule assisting in the differentiation of complex parts within the products. Whenever dealing with the multiplication of functions in calculus, remembering the product rule is invaluable.

Derivatives: Fundamental Calculus Concept

Derivatives are the core of calculus, representing the rate of change of a function. If you imagine a curve expressed by a function, the derivative at a point gives the slope of the tangent to that curve at the same point. In simpler terms, it quantifies how one variable changes in response to changes in another.In the realm of implicit differentiation, like in the given exercise, derivatives allow us to understand how variables related in complex ways change simultaneously. We often find situations where explicitly solving for one variable in terms of another is cumbersome or impossible. In such cases, we can distinguish both sides of the equation with respect to \( x \), while treating \( y \) as an implicit function of \( x \).For this reason, we use the implicit differentiation technique to find \( \frac{dy}{dx} \). After applying the chain and product rules in our differentiation process, derivatives from both sides of the original equation brought us to the simplified expression. This understanding shows the strategic use of derivatives makes complex relationships much simpler to decipher and solve.