Problem 11 - Calculus of a Single Variable

Question

Find \(d y / d x\) by implicit differentiation. \(\sin x+2 \cos 2 y=1\)

Step-by-Step Solution

Answer

The derivative of \(y\) with respect to \(x\) by implicit differentiation of the given function is \(y' = \frac{\cos x}{4\sin 2y}\)

Step 1: Differentiate the equation with respect to \(x\)

Differentiate each term of the implicit equation. The derivative of \(\sin x\) with respect to \(x\) is \(\cos x\). The term \(2 \cos 2y\) is differentiated as a chain rule: The derivative of \(\cos u\) with respect to \(u\) is \(-\sin u\), hence the derivative of \(2 \cos 2y\) is \(-2 \sin (2y) \cdot 2 dy/dx\). The derivative of a constant \(1\) is \(0\) as it does not change. So the differentiated equation is: \(\cos x - 4y'\sin 2y = 0\) where \(y' = dy/dx\).

Step 2: Isolate \(y'\)

Now, rearrange the differentiated equation for \(y'\): \(y' = \frac{\cos x}{4\sin 2y}\).

Key Concepts

Derivatives

In calculus, derivatives are a fundamental concept used to determine the rate at which one quantity changes with respect to another. When dealing with functions, the derivative at a certain point gives us the slope of the tangent line at that point, which essentially describes how steep the function is at that point. The process of finding a derivative is called differentiation. When differentiating simple algebraic expressions, the rules are quite straightforward. However, when involving more complex expressions such as trigonometric functions or when the relationship in the equation is not explicit, it requires more sophisticated methods such as implicit differentiation. Implicit differentiation becomes handy when we need to find the derivative of expressions where the dependent and independent variables are intertwined, allowing us to solve for derivatives without isolating the variable initially.

Chain Rule

The chain rule is a powerful technique in calculus used to differentiate composite functions. A composite function is essentially a function that is made up of several smaller functions. If you imagine the process of solving a composite function like peeling an onion, the chain rule helps you peel back each layer systematically, differentiating one layer at a time.The rule states that to differentiate a composite function, you differentiate the outer function, and then multiply by the derivative of the inner function. For example, in the exercise with the term \(2 \cos 2y\), we apply the chain rule as follows:

• Differentiated outer function: The derivative of \(\cos u\) is \(-\sin u\).
• Internal derivative: The derivative of \(2y\) with respect to \(x\) is \(2 \frac{dy}{dx}\).

This way, we find the derivative correctly as \(-4 \sin(2y) \cdot \frac{dy}{dx}\). This ensures all components are addressed, revealing how changes in each part affect the overall function.

Trigonometric Functions

Trigonometric functions are essential functions in calculus that relate angles of a triangle to the lengths of its sides. They include the sine, cosine, and tangent functions, amongst others. These functions are periodic, meaning they repeat values in regular intervals, which makes them unique and often more complex to differentiate.In our exercise, trigonometric functions like \(\sin x\) and \(\cos 2y\) are involved. When differentiating trigonometric functions:

• The derivative of \(\sin x\) is \(\cos x\).
• The derivative of \(\cos x\) is \(-\sin x\).

Combining trigonometric differentiation with techniques like the chain rule aids in handling more complex trigonometric expressions smoothly. Remember, understanding these derivatives is key in calculus problems involving periodic behavior or wave patterns, where trigonometric functions are predominantly used.

Calculus Problem Solving

Solving calculus problems, especially involving derivatives, is all about methodical and logical thinking. It requires breaking down a problem into more manageable parts and applying the appropriate calculus rules to each part. For instance, the given problem requires us to use implicit differentiation, triggered by the mixture of trigonometric functions and implicitly related variables such as \(x\) and \(y\). The sequence of steps includes:

• Identifying the differentiable components (like \(\sin x\) and \(2 \cos 2y\)).
• Applying differentiation rules like the power rule, product rule, and particularly implicit and the chain rule.
• Rearranging the terms to solve for \(\frac{dy}{dx}\), isolating the desired derivative.

Each step in solving these problems builds on critical thinking and a clear understanding of calculus principles, ensuring any approach or method remains consistent and correct throughout the exercise.