Problem 65 - Calculus - AP Edition

Question

Find \(d y / d x\) by implicit differentiation. $$x^{3}+x \tan ^{-1} y=e^{y}$$

Step-by-Step Solution

Answer

\(\frac{dy}{dx} = \frac{-(3x^2 + \tan^{-1}(y))}{x - e^y}\)

Step 1: Differentiate Both Sides

Apply implicit differentiation to both sides of the equation: \(x^3 + x\tan^{-1}(y) = e^y\). Differentiate term-by-term with respect to \(x\), remembering to use the chain rule for terms involving \(y\).

Step 2: Differentiate the Left Side

For \(x^3\), use the power rule to get \(3x^2\). For \(x\tan^{-1}(y)\), apply the product rule: differentiate \(x\) to get \(1\) and \(\tan^{-1}(y)\) differentiates to \(\frac{1}{1+y^2} \cdot \frac{dy}{dx}\), resulting in \(\tan^{-1}(y) + x \frac{1}{1+y^2} \frac{dy}{dx}\).

Step 3: Differentiate the Right Side

For \(e^y\), use the chain rule: the derivative is \(e^y \cdot \frac{dy}{dx}\).

Step 4: Combine All Derivatives

Combine all differentiated terms: \[3x^2 + \tan^{-1}(y) + x \frac{1}{1+y^2} \frac{dy}{dx} = e^y \cdot \frac{dy}{dx}\]

Step 5: Isolate \(\frac{dy}{dx}\)

Rearrange the equation to solve for \(\frac{dy}{dx}\). Move terms without \(\frac{dy}{dx}\) to the other side:\[x \frac{dy}{dx} - e^y \frac{dy}{dx} = - (3x^2 + \tan^{-1}(y))\]Factor out \(\frac{dy}{dx}\):\[\frac{dy}{dx}(x - e^y) = - (3x^2 + \tan^{-1}(y))\]

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

Divide both sides by \((x - e^y)\) to isolate \(\frac{dy}{dx}\):\[\frac{dy}{dx} = \frac{-(3x^2 + \tan^{-1}(y))}{x - e^y}\]

Key Concepts

Chain Rule

The chain rule is a fundamental concept in differentiation techniques, used to differentiate compositions of functions. When dealing with an equation where functions depend on another variable, the chain rule allows us to account for these dependences step-by-step. In implicit differentiation, the chain rule is particularly useful because we often need to differentiate compositions involving both given functions and implicit functions.
The general idea of the chain rule is to differentiate the outer function with respect to its inner function, and then multiply by the derivative of the inner function. For example, if we have a function like \(f(g(x))\), the derivative is \(f'(g(x)) \cdot g'(x)\).

• When differentiating \(x^3 + x \tan^{-1}(y) = e^y\) implicitly, use the chain rule for terms like \(\tan^{-1}(y)\) and \(e^y\) since \(y\) is a function of \(x\).
• For \(\tan^{-1}(y)\), recognize it as a composition \(\tan^{-1}(y(x))\), leading to \(\frac{1}{1+y^2} \cdot \frac{dy}{dx}\).
• Similarly, \(e^y\) differentiates as \(e^y \cdot \frac{dy}{dx}\).

Understanding the chain rule is essential as it helps tackle more complex differentiation problems efficiently.

Product Rule

The product rule is essential when dealing with the differentiation of products of functions. It states that if you have two functions merged through multiplication, like \(u(x)\) and \(v(x)\), their derivative is obtained through \[u'(x)v(x) + u(x)v'(x)\].When addressing the problem \(x\tan^{-1}(y)\), you see a perfect opportunity to apply the product rule:

• Identify \(u(x) = x\) and \(v(x) = \tan^{-1}(y)\).
• Differentiating \(x\) gives \(1\), and differentiating \(\tan^{-1}(y)\) with respect to \(x\) uses the chain rule, resulting in \(\frac{1}{1+y^2} \cdot \frac{dy}{dx}\).
• Combine these using the product rule: \(1 \cdot \tan^{-1}(y) + x \cdot \frac{1}{1+y^2} \frac{dy}{dx}\).

This approach ensures every part of the expression is correctly accounted for, a crucial step in calculus problem-solving.

Differentiation Techniques

Differentiation techniques are the strategies and methods you implement to find the derivative of functions effectively. In the given problem, multiple techniques like the chain rule, product rule, and implicit differentiation work together.
These techniques allow you to handle a variety of expressions in calculus:

• **Implicit Differentiation:** Used when functions are not expressed explicitly as \(y = f(x)\) but are tangled in an equation. Here, you differentiate both sides with respect to \(x\) while applying the rules for each term.
• **Power Rule:** Simply states if you have \(x^n\), the derivative is \(nx^{n-1}\). Straightforward and used on terms like \(x^3\).
• By using these methods strategically, you can derive solutions for complex equations step-by-step.

Mastering these techniques enhances your calculus problem-solving abilities significantly.

Calculus Problem Solving

Problem-solving in calculus involves breaking down complex mathematical expressions into manageable parts, applying differentiation techniques systematically. Implicit differentiation often comes into play with these problems, forming a crucial part of the solution process.
The goal is to find \(\frac{dy}{dx}\), the derivative of \(y\) with respect to \(x\), often within equations where \(y\) is not isolated:

• Start by differentiating every component in the equation using appropriate techniques, like the chain or product rule, and gather all derivatives.
• Next, collect terms containing \(\frac{dy}{dx}\) and isolate them on one side of the equation.
• Solve this simplified expression algebraically to find \(\frac{dy}{dx}\).

This structured approach to solving calculus problems teaches you not just to find a solution, but to understand the relationships and dependencies within the equation. Each step logically builds on the last, exemplifying how calculus aids in exploring and solving real-world problems.