Question

Find \(d y / d x\). (Treat \(a\) and \(r\) as constants.) $$x y=5$$

Short answer

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

Step-by-step solution

Differentiate both sides with respect to x

To find the derivative of the given implicit function with respect to x, apply the product rule which states that the derivative of a product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function. In this case, use the product rule to differentiate the left side, and because the right side is a constant, its derivative will be zero.

Apply the product rule

The product rule is given by \(d(uv)/dx = u'v + uv'\). Here, apply the product rule where \(u = x\) and \(v = y\). Thus, the derivative of \(xy\) is \(x(dy/dx) + y(1)\).

Set derivative of constant to zero

The derivative of a constant, in this case, 5, is zero. So the right-hand side of the equation becomes zero when differentiated.

Solve for \(dy/dx\)

Now solve the resulting equation for \(dy/dx\). Combine terms involving \(dy/dx\) on one side to isolate it.

Key concepts

Derivative of an Implicit Function

Understanding the derivative of an implicit function is essential in calculus, especially when dealing with equations where it is not feasible or possible to solve for one variable in terms of another. Implicit differentiation is a technique that aids in finding the derivative of a function that is not explicitly solved for one variable.

In the given exercise with the equation \(xy = 5\), \(y\) is not expressed explicitly as a function of \(x\); thus, we need to apply implicit differentiation. Start by differentiating both sides of the equation with respect to \(x\), treating \(y\) as a function of \(x\). When differentiating the product of \(\(x\) and \(\)y\), we apply the product rule, which will be discussed separately.

After applying the product rule, terms containing \(dy/dx\) appear. In order to find the value of \(dy/dx\), we gather these terms on one side of the equation and solve. This process provides the derivative of \(y\) with respect to \(x\) even though \(y\) has not been given in terms of \(x\). To further solidify this concept, practice with various implicit functions is advisable, as this improves computational flexibility and deepens the understanding of how derivatives interplay with algebraic manipulation.

Product Rule in Calculus

The product rule is a fundamental rule in calculus for finding the derivative of a product of two functions. It's formally stated as: if you have two differentiable functions \(u(x)\) and \(v(x)\), their derivative with respect to \(x\) is given by \(d(uv)/dx = u'v + uv'\). This rule is essential when functions are intertwined by multiplication, as in our exercise with the function \(xy\).

Application in the Exercise

When we differentiate \(xy\), we view \(x\) as \(u\) and \(y\) as \(v\). By the product rule, we then have \(x(dy/dx) + y(1)\) because the derivative of \(x\) with respect to \(x\) is 1 and \(y\) is treated as a function of \(x\) in this context. It's imperative to understand and apply the product rule correctly, as it often serves as a stepping stone towards mastery in techniques of differentiation and is widely applicable in various problems.

Differentiating Constants

Differentiating constants is one of the simpler yet profoundly important concepts in calculus. The main rule to remember is that the derivative of a constant with respect to any variable is zero. This rule extends from the fact that a constant does not change at all — it has no rate of change, hence the derivative, which represents the rate of change, is zero.

Role in the Provided Exercise

In our exercise, the constant is the number 5 on the right side of the equation \(xy = 5\). According to our rules of differentiation, when we take the derivative of both sides of the equation with respect to \(x\), the derivative of 5 is 0. That's why, after applying the product rule to the left-hand side, we set the right-hand side of the equation equal to zero. This principle is especially handy when simplifying derivatives and is broadly applicable whenever constants are present in functions being differentiated. Recognizing and applying this concept correctly streamlines the process and avoids common pitfalls in calculus problems.