Problem 30 - Thomas Calculus

Question

Use implicit differentiation to find \(d y / d x\) in Exercises \(19-32\) $$ x+\sin y=x y $$

Step-by-Step Solution

Answer

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

Step 1: Rearrange the Equation

We start with the given equation: \[ x + \sin y = x y \] which can be rearranged as:\[ x + \sin y - xy = 0 \] This form is useful for applying implicit differentiation as everything is set to zero.

Step 2: Differentiate with Respect to x

Differentiate both sides of the equation with respect to \( x \). Remember to apply the chain rule when differentiating \( \sin y \) as it is a function of \( y \), and use the product rule for differentiating \( xy \). \[ \frac{d}{dx}(x) + \frac{d}{dx}(\sin y) - \frac{d}{dx}(xy) = 0 \] This simplifies to: \[ 1 + \cos y \cdot \frac{dy}{dx} - (y + x \cdot \frac{dy}{dx}) = 0 \]

Step 3: Rearrange to Isolate \( \frac{dy}{dx} \)

Rearrange the differentiated equation to solve for \( \frac{dy}{dx} \):\[ 1 + \cos y \cdot \frac{dy}{dx} - y - x \cdot \frac{dy}{dx} = 0 \] Combine like terms involving \( \frac{dy}{dx} \): \[ (\cos y - x) \cdot \frac{dy}{dx} = y - 1 \]

Step 4: Solve for \( \frac{dy}{dx} \)

Finally, solve for \( \frac{dy}{dx} \) by dividing both sides of the equation by \((\cos y - x)\):\[ \frac{dy}{dx} = \frac{y - 1}{\cos y - x} \] This expression gives us the derivative \( \frac{dy}{dx} \) using implicit differentiation.

Key Concepts

calculus problem-solving

Solving calculus problems often involves understanding and applying several mathematical techniques. One common challenge is implicit differentiation, which is particularly useful when dealing with equations that are not explicitly solved for one variable in terms of another. In the given exercise, we have an equation involving both \(x\) and \(y\), but \(y\) is not isolated on one side. This implies that we must consider how these variables interact throughout the problem. The key steps in a calculus problem like this include:

• Rearranging the equation so that it equals zero, providing a clear starting point for applying differentiation techniques.
• Using differentiation rules mindfully, such as the chain rule and the product rule, to tackle different parts of the equation.
• Rearranging the differentiated equation to solve for the derivative of the interest, \(\frac{dy}{dx}\).

The methodical approach not only simplifies a tangled relationship between variables but also clarifies each step needed to reach the solution. Using these strategic steps, solutions become more accessible and less daunting.

derivatives

Derivatives play a crucial role in calculus, offering insights into the rate at which one quantity changes with respect to another. In problems that require implicit differentiation, derivatives help us understand how changes in \(x\) affect \(y\), even when \(y\) is implicitly defined. When we differentiate the equation:\[ x + \sin y - xy = 0 \]we do so with respect to \(x\), treating \(y\) as a function of \(x\). This process enables us to explore how the presence of \(x\) influences \(y\), revealing the interconnected nature of their relationship.
In general, understanding derivatives and their applications provides powerful tools for modeling and predicting real-world phenomena, from physics to economics.

chain rule

The chain rule is a fundamental technique in differential calculus used when differentiating composite functions. It helps us differentiate expressions where one function is nested within another. In our problem, we encounter the \(\sin y\) term, which requires the application of the chain rule because \(y\) itself is a function of \(x\).Let's break it down:

• The outer function here is \(\sin\) and the inner function is \(y\).
• When differentiating \(\sin y\) with respect to \(x\), the chain rule tells us to first differentiate the outer function, \(\sin y\), with respect to \(y\), giving \(\cos y\).
• Then, multiply that result by the derivative of the inner function, \(\frac{dy}{dx}\).

This results in \(\cos y \cdot \frac{dy}{dx}\). The chain rule's systematic application ensures we correctly account for all layers of dependency between variables, vital for correct solutions in implicit differentiation.

product rule

The product rule is essential when differentiating products of two functions. In implicit differentiation, it's often applied when terms in the equation are products of \(x\) and \(y\), as seen in the \(xy\) term of our exercise equation.The product rule states:

• If you have two functions \(u(x)\) and \(v(x)\), then their product's derivative is \(u'(x) v(x) + u(x) v'(x)\).

For our term \(xy\), \(u(x) = x\) and \(v(x) = y\), it implies:

• \(u'(x) = 1\) because the derivative of \(x\) is 1, and \(v'(x) = \frac{dy}{dx}\) because \(y\) is a function of \(x\).
• Applying the product rule, we get \(1 \cdot y + x \cdot \frac{dy}{dx}\).

Remember, the product rule allows us to disentangle complex products in differentiation, yielding a clear path to solving implicit differentiation problems.