Problem 31 - Thomas Calculus

Question

find \(d y / d x.\) \begin{equation}3+\sin y=y-x^{3}\end{equation}

Step-by-Step Solution

Answer

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

Step 1: Move All Terms to One Side

First, rearrange the equation to gather all terms on one side: \( y - \sin(y) + x^3 - 3 = 0 \). This form will make differentiation easier.

Step 2: Differentiate with Respect to x

Differentiate each term with respect to \(x\). Using implicit differentiation:- The derivative of \(y\) with respect to \(x\) is \(\frac{dy}{dx}\).- The derivative of \(-\sin(y)\) is \(-\cos(y) \cdot \frac{dy}{dx}\) (by chain rule).- The derivative of \(x^3\) is \(3x^2\).This gives: \(\frac{dy}{dx} - \cos(y) \cdot \frac{dy}{dx} + 3x^2 = 0\).

Step 3: Factor Out dy/dx

Factor \(\frac{dy}{dx}\) out from the terms where it appears: \(\frac{dy}{dx}(1 - \cos(y)) = -3x^2\).

Step 4: Solve for dy/dx

To solve for \(\frac{dy}{dx}\), divide both sides by \((1 - \cos(y))\): \[ \frac{dy}{dx} = \frac{-3x^2}{1 - \cos(y)} \].

Key Concepts

Understanding the Chain Rule

The chain rule is a fundamental tool in calculus, particularly useful for differentiating composite functions. Imagine you have a function nested within another function, similar to the layers of an onion. For example, in our original problem with terms like \(-\sin(y)\), the composition of functions is important. Here, \(-\sin(y)\) is a function of \(y\), which itself is a function of \(x\).

• To differentiate such compositions, the chain rule helps by instructing you to take the derivative of the outer function while keeping the inner function intact.
• Then, you multiply the result by the derivative of the inner function.

In our exercise, the outer function is \(\sin(y)\), whose derivative is \(\cos(y)\). The inner function is \(y\), which when differentiated with respect to \(x\) gives \(\frac{dy}{dx}\).
By applying the chain rule, the derivative of \(-\sin(y)\) becomes \(-\cos(y) \cdot \frac{dy}{dx}\). This concept is a cornerstone in implicit differentiation and highlights the interconnectedness of variables.

Application of Differential Calculus

Differential calculus is the branch of mathematics focused on studying how functions change. In the context of our exercise, it is about finding the rate at which \(y\) changes with respect to \(x\), expressed as \(\frac{dy}{dx}\).
This is crucial because direct differentiation is not always possible, especially when dealing with implicit functions like the given equation. By rearranging the equation to bring all terms to one side, \(y - \sin(y) + x^3 - 3 = 0\), we expose the structure for differentiation.

• Differential calculus techniques allow us to systematically find \(\frac{dy}{dx}\) by differentiating each term with respect to \(x\).
• This includes treating some terms, such as \(y\), as implicit functions of \(x\).
• We use standard derivatives for simple terms (like \(x^3\) which becomes \(3x^2\)), while composite terms invoke the chain rule.

The implicit differentiation approach forces us to carefully consider the relationships between variables, aiding in solving more complex equations that cannot be easily isolated.

Solving the Equation for dy/dx

Once we have differentiated all terms in our implicit equation, the goal is to isolate \(\frac{dy}{dx}\). Solving equations of this nature often involves balancing both sides until the desired variable stands on its own.
In the given problem, after differentiating, we acquired this equation:\[\frac{dy}{dx} - \cos(y) \cdot \frac{dy}{dx} + 3x^2 = 0\]To solve for \(\frac{dy}{dx}\), follow these steps:

• Notice that \(\frac{dy}{dx}\) is a common factor in two terms. Factor out \(\frac{dy}{dx}\).
• Rearranged, the equation becomes: \( \frac{dy}{dx}(1 - \cos(y)) = -3x^2 \).
• Then, isolate \(\frac{dy}{dx}\) by dividing both sides by \((1 - \cos(y))\).
• The solution: \( \frac{dy}{dx} = \frac{-3x^2}{1 - \cos(y)} \).

This method emphasizes the importance of algebraic manipulation in solving equations, particularly those derived from implicit differentiation. This approach not only provides the solution but deepens understanding of functional relationships in calculus.