Problem 33 - Calculus Early Transcendentals

Question

Given the following equations, evaluate \(d y / d x .\) Assume that each equation implicitly defines \(y\) as a differentiable function of \(x\). $$2 \sin x y=1$$

Step-by-Step Solution

Answer

Answer: The derivative of y with respect to x is 0, i.e., dy/dx = 0.

Step 1: Differentiate both sides of the equation with respect to x

Start by differentiating both sides of the given equation \(2 \sin x y = 1\) with respect to \(x\). Remember that when differentiating a product of two functions, we must use the product rule: \((uv)' = u'v + uv'\). Applying this rule and the chain rule to the equation, we get: $$\frac{d}{dx}(2 \sin x y) = \frac{d}{dx}(1)$$

Step 2: Apply the product rule and chain rule

Now let's apply the product rule and chain rule. On the left side of the equation, the derivative of \(2 \sin x\) with respect to \(x\) is \(2 \cos x\), and the derivative of \(y\) with respect to x is \(\frac{dy}{dx}\). So we have: $$2 \cos x \frac{dy}{dx} + 2 \sin x \frac{d y}{dx} = 0$$ On the right side of the equation, the derivative of a constant (1) with respect to x is 0. So, the equation becomes: $$2 \cos x \frac{dy}{dx} + 2 \sin x y' = 0$$

Step 3: Solve for dy/dx

Now we can isolate \(\frac{dy}{dx}\) by factoring it out and dividing both sides by \((2 \cos x + 2 \sin x)\): $$\frac{dy}{dx}(2 \cos x + 2 \sin x) = 0$$ Dividing by \((2 \cos x + 2 \sin x)\): $$\frac{dy}{dx} = \frac{0}{2 \cos x + 2 \sin x}$$ Therefore, $$\frac{dy}{dx} = 0$$ This means the derivative of \(y\) with respect to \(x\) is 0 i.e., \(y\) is a constant function.

Key Concepts

Product Rule

The Product Rule is an essential derivative rule applied when differentiating expressions that involve products of two or more functions. In such cases, simply differentiating each function individually and multiplying them together is incorrect. Instead, the Product Rule provides a more systematic approach.

• The Formula: For two functions, say \( u(x) \) and \( v(x) \), their derivative according to the Product Rule is \((uv)' = u'v + uv' \).


• Practical Interpretation: When you differentiate \( u \) and multiply by \( v \), and vice versa, ensuring to add both results together. This considers all aspects of how both functions and their interactions change.

In our specific problem, the equation is \( 2 \sin(xy) = 1 \). To solve this, we see there is a multiplication of \( 2 \sin x \) and \( y \). Applying the Product Rule here helps systematically manage the differentiation of such complexities in implicit functions.

Chain Rule

The Chain Rule is a powerful tool in calculus when dealing with composite functions, meaning functions nested within each other like \( f(g(x)) \). It helps us find the derivative of a function that is "inside" another function.

• The Formula: The derivative of \( f(g(x)) \) is \( f'(g(x)) \cdot g'(x) \). Essentially, differentiate the outer function while keeping the inner function constant, and multiply by the derivative of the inner function.


• Why It Matters: The Chain Rule is crucial because it allows us to handle complex situations where one function’s output is another's input, a common scenario in implicit differentiation where functions of \(x\) and \(y\) are intertwined.

In our exercise, when differentiating \( 2\sin(xy) \), the Chain Rule helps tackle \( \sin(xy) \), treating \( xy \) as the "inner" function, thus allowing proper integration with the Product Rule to find the derivative successfully.

Differentiable Functions

Differentiable functions are at the heart of calculus, denoting functions that have derivatives at every point in their domain. This means that the function is smooth and without jumps or breaks, ensuring that the derivative exists everywhere along it.

• Definition: A function \( f \) is differentiable at \( x_0 \) if the derivative \( f'(x_0) \) exists. This also implies the function is continuous at that point.


• Importance in Implicit Differentiation: When a problem states that \( y \) is a differentiable function of \( x \), it means you can apply differentiation techniques across all relevant portions of \( y \) without concern for undefined behavior.

In the exercise, assuming \( y \) is a differentiable function of \( x \) is crucial because it allows using implicit differentiation techniques to solve for \( \frac{dy}{dx} \). Without differentiability, such operations either become extremely complex or impossible to achieve reliably.