Problem 10 - 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. \(x \sqrt{y+1}=x y+1\)

Step-by-Step Solution

Answer

\(y' = \frac{y - \sqrt{y+1}}{\frac{x}{2\sqrt{y+1}} - x}\).

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

Start by differentiating the given equation, \(x \sqrt{y+1} = xy + 1\), with respect to \(x\). Use the product rule and chain rule where necessary.

Step 2: Differentiate the left side of the equation

For the left side, \(x \sqrt{y+1}\), apply the product rule: \([f(x)g(x)]' = f'(x)g(x) + f(x)g'(x)\). Here, \(f(x) = x\) and \(g(x) = \sqrt{y+1}\). We get \(1 \cdot \sqrt{y+1} + x \cdot \frac{1}{2\sqrt{y+1}} \cdot (y') = \sqrt{y+1} + \frac{xy'}{2\sqrt{y+1}}\).

Step 3: Differentiate the right side of the equation

For the right side, \(xy + 1\), differentiate term by term: \((xy)' = x y' + y\) (using the product rule), and the derivative of the constant \(1\) is \(0\). Thus, \(x y' + y\).

Step 4: Set the derivatives equal and solve for y'

Equate the derivatives from the left and right sides: \(\sqrt{y+1} + \frac{xy'}{2\sqrt{y+1}} = x y' + y\). Rearrange the terms to solve for \(y'\): \(\frac{xy'}{2\sqrt{y+1}} - x y' = y - \sqrt{y+1}\).

Step 5: Factor and solve for y'

Factor out \(y'\) from the terms on the left: \(y'\left(\frac{x}{2\sqrt{y+1}} - x\right) = y - \sqrt{y+1}\). Solve for \(y'\) by isolating it: \(y' = \frac{y - \sqrt{y+1}}{\frac{x}{2\sqrt{y+1}} - x}\). Simplify this expression further if necessary.

Key Concepts

Product Rule

The product rule is a fundamental tool in calculus, essential for differentiating expressions where two functions are multiplied together. Imagine you have an equation featuring the product of two functions, say \( f(x) \) and \( g(x) \). The product rule helps you find the derivative of this product. The formula is:

• \((f(x)g(x))' = f'(x)g(x) + f(x)g'(x)\)

In simpler terms, you differentiate the first function, multiply it by the second function, then differentiate the second function, and multiply it by the original first function. Add those two results together.
To see it in action, take the term \( x\sqrt{y+1} \). We identify \( f(x) = x \) and \( g(x) = \sqrt{y+1} \). Applying the product rule:

• Differentiate \( x \) to get \( 1 \), keep \( \sqrt{y+1} \) as is: \( 1 \cdot \sqrt{y+1} \).
• Now, keep \( x \) as is and differentiate \( \sqrt{y+1} \), taking care to use the chain rule for \( \sqrt{y+1} \). This yields \( x \cdot \frac{1}{2\sqrt{y+1}} \cdot y' \).

Putting these together, the derivative of \( x\sqrt{y+1} \) using the product rule is \( \sqrt{y+1} + \frac{xy'}{2\sqrt{y+1}} \). This showcases how the product rule enables untangling of compounded differentiations.

Chain Rule

The chain rule is a powerful differentiation technique used whenever you're dealing with compositions of functions. Imagine you have a function nested within another, like \( y = \sqrt{y+1} \). The chain rule helps you differentiate such functions neatly.
The basic idea is to take the derivative of the outer function while preserving the inner function as it is, then multiply by the derivative of the inner function.

• The rule can be written as: If \( y = g(f(x)) \), then \( y' = g'(f(x)) \cdot f'(x) \).

In our example, \( y = \sqrt{y+1} \), the outer function is \( g(u) = \sqrt{u} \) where \( u = y+1 \), and the inner function is \( u = y+1 \).
Differentiate \( \sqrt{u} \) to get \( \frac{1}{2\sqrt{u}} \) and \( u = y+1 \) provides the derivative \( y' \), combining them yields \( \frac{1}{2\sqrt{y+1}} \cdot y' \).
Applying the chain rule allows us to break down the derivative of more complex expressions into manageable steps, simplifying the learning process.

Differentiable Functions

Differentiable functions are smooth and continuous, meaning they have a defined slope at every point along their path. This concept is critical for using calculus effectively.
A differentiable function, by definition, has a derivative that exists everywhere in its domain — a key characteristic for the analyses we perform in mathematics.

• For example, basic polynomials, exponentials, trigonometric functions, and even more complex compositions generally fall into the category of differentiable functions.
• In calculus, establishing that a function is differentiable is the gateway to using rules like the product and chain rules.
• Functions with sharp corners or discontinuities, like the absolute value function \( |x| \), can often mark spots where a function is not differentiable.

Differentiability assures us the function is well-behaved enough to be examined using the powerful tools of calculus, allowing us to derive insights on the behavior and properties of such functions with confidence.