Problem 11 - Calculus Early Transcendentals: Pearson New International Edition

Question

Assuming that each equation defines a differentiable function of \(x\), find \(D_{x} y\) by implicit differentiation. \(x y+\sin (x y)=1\)

Step-by-Step Solution

Answer

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

Step 1: Differentiate Both Sides of the Equation

We begin with the equation \( x y + \sin(x y) = 1 \). To find \( D_x y \) using implicit differentiation, differentiate both sides of the equation with respect to \( x \). When differentiating, remember to use the product rule for \( xy \) and the chain rule for \( \sin(xy) \).

Step 2: Apply the Product and Chain Rules

The derivative of \( xy \) with respect to \( x \) is \( y + x \frac{dy}{dx} \). The chain rule gives us the derivative of \( \sin(xy) \) as \( \cos(xy)(y + x \frac{dy}{dx}) \). Therefore, the differentiation yields:\[ y + x \frac{dy}{dx} + \cos(xy)(y + x \frac{dy}{dx}) = 0 \].

Step 3: Simplify the Expression

Now, simplify the expression:\[ y + x \frac{dy}{dx} + y \cos(xy) + x \cos(xy) \frac{dy}{dx} = 0 \]Combine the terms with \( \frac{dy}{dx} \):\[ (x + x \cos(xy)) \frac{dy}{dx} = -y - y \cos(xy) \].

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

Isolate \( \frac{dy}{dx} \) by dividing both sides of the equation by \( x + x \cos(xy) \):\[ \frac{dy}{dx} = \frac{-y(1 + \cos(xy))}{x(1 + \cos(xy))} \].Since \( 1 + \cos(xy) \) can cancel out (assuming it's not zero), we get:\[ \frac{dy}{dx} = \frac{-y}{x} \].

Key Concepts

Product Rule

When dealing with the term \( xy \) in the equation \( xy + \sin(xy) = 1 \), we need to apply the product rule. The product rule is a simple but crucial differentiation technique used when you have to differentiate products of two functions. It states:

• If you have two functions \( u(x) \) and \( v(x) \), the derivative of their product with respect to \( x \) is given by \( u'(x)v(x) + u(x)v'(x) \).

Here, \( u(x) = x \) and \( v(x) = y \). Thus, when we differentiate \( xy \) with respect to \( x \), we treat \( y \) as a function of \( x \), and use the rule:

• \( \, [x \, y]' = x \frac{dy}{dx} + y \) ,

where \( y \frac{dy}{dx} \) comes from differentiating \( y \) with respect to \( x \). Remember, the product rule is essential when multiple variables depend on each other, and you differentiate them concerning one variable.

Chain Rule

Next up is the chain rule, which plays a crucial role in differentiating the \( \sin(xy) \) part of our equation. The chain rule is essential for differentiating composite functions—functions within functions. It's given by:

• If you have a composite function \( f(g(x)) \), then the derivative is \( f'(g(x)) \cdot g'(x) \).

For \( \sin(xy) \), think of it as \( f(u) = \sin(u) \), where \( u = xy \). Therefore, the derivative is:

• First, differentiate \( \sin(u) \) to get \( \cos(u) \).
• Then, multiply by the derivative of \( u \), which is \( \frac{d}{dx}(xy) = y + x \frac{dy}{dx} \), thanks to our product rule!

So, the chain rule gives the derivative of \( \sin(xy) \) as \( \cos(xy)(y + x \frac{dy}{dx}) \). The chain rule bridges layers of functions, letting us find derivatives of complex nested functions.

Differentiable Function

In the context of finding \( D_x y \) through implicit differentiation, it's vital to understand what differentiable functions are. A function is differentiable at a point if it has a derivative there, meaning there exists a tangent at every point in its domain.

• For instance, classic differentiable functions include all polynomials, exponential, sine, and cosine functions, among others.
• To be differentiable, a function must be continuous—meaning no breaks, jumps, or sharp points graphically—and smooth.

In the exercise, the assumption is that each equation defines a differentiable function of \(x\), which allows us to use implicit differentiation confidently.
Implicit differentiation is an approach where you differentiate an equation involving x and y with respect to x without solving for y first. It applies when y is a hidden function of x, shedding light on how y changes as x changes. Through the process of differentiating, using the product and chain rules, we found how to derive \( \frac{dy}{dx} \) for a hidden relationship between x and y.