Question

In Exercises \(25-34,\) find \(d y / d x\) by implicit differentiation and evaluate the derivative at the indicated point. $$ x y=4, \quad(-4,-1) $$

Short answer

The derivative \(\frac{d y}{d x}\) of the function at the point (-4, -1) is -1/4.

Step-by-step solution

Apply Product Rule

The product rule states that the derivative of two functions multiplied together is the first function times the derivative of the second function, plus the second function times the derivative of the first function. Apply this rule to \(x y = 4\) to get \(x \frac{d y}{d x} + y = 0\).

Solve For \(\frac{d y}{d x}\)

To isolate \(\frac{d y}{d x}\), rearrange the equation obtained in step 1: \(\frac{d y}{d x} = -\frac{y}{x}\).

Evaluate the Derivative at the Given Point

Substitute the given point, (-4, -1), into the derivative equation to find the value of \(\frac{d y}{d x}\) at that point: \(\frac{d y}{d x} = -\frac{-1}{-4} = -\frac{1}{4}\).

Key concepts

Product Rule

Understanding the product rule is essential when dealing with the differentiation of products of two functions. In calculus, the product rule is a way to find the derivative of a function that is the product of two other functions which we'll denote here as u(x) and v(x). It states that the derivative of the product u(x)v(x) is given by u'(x)v(x) + u(x)v'(x), where u'(x) and v'(x) are the derivatives of u(x) and v(x) respectively.

The significance of the product rule can be seen in a variety of calculus exercises. When faced with a product of functions, rather than expanding the product which can sometimes be cumbersome or impossible, we can use the product rule for a more straightforward differentiation process. For example, when given a function like the one in our exercise, xy, you can directly apply the product rule to find the derivative with respect to x, leading us to the first step towards solving the problem.

Derivative Evaluation

Once the derivative of an implicitly defined function is found, we often want to evaluate the derivative at a specific point. This step is called derivative evaluation. To effectively perform a derivative evaluation, it's critical to understand that after taking the derivative, the resulting expression can then be treated as a function of x, which can be evaluated at specific values of x and y.

Evaluating the derivative at a particular point provides us with the slope of the tangent line at that point on the graph of the function. In our exercise, after applying the product rule, we obtain an expression for \(\frac{dy}{dx}\) that includes variables x and y. By substituting the coordinates of the given point into this expression, we calculate the numerical value of the derivative at that point, which can be seen in the final step of the provided solution.

Calculus Exercises

Solving calculus exercises like implicit differentiation problems enhances understanding and application of derivative concepts in various functions. Implicit differentiation is particularly useful when dealing with equations that define y implicitly rather than explicitly as a function of x. Here, instead of isolating y and differentiating, we differentiate both sides of the equation with respect to x directly and solve for \(\frac{dy}{dx}\), keeping in mind to use the chain rule for terms involving y.

As seen in this exercise, the process of taking the derivative of both sides of an equation with respect to x may involve the product rule, quotient rule, or chain rule, depending on the form of the equation provided. Regular practice of these exercises helps not only to grasp the mechanics of differentiation but also its applications when equations are not straightforward to solve explicitly. It is a fundamental skill that aids in problem-solving across different areas of calculus.