Problem 12 - College Algebra and Calculus: An Applied Approach

Question

Find \(d y / d u, d u / d x\), and \(d y / d x\). $$ y=2 \sqrt{u}, u=5 x+9 $$

Step-by-Step Solution

Answer

The derivatives are \(d y / d u = 1/\sqrt{u}\), \( d u / d x = 5 \) and \(d y / d x = 5/\sqrt{5x+9}\).

Step 1: Derivative of y with respect to u

The first part is to find \(d y / d u\). The original function given is \(y=2\sqrt{u}\). The rule for derivative of \(\sqrt{u}\) with respect to \(u\) is \(0.5/\sqrt{u}\). So, \(d y / d u = 2*0.5/\sqrt{u} = 1/\sqrt{u}\).

Step 2: Derivative of u with respect to x

The next step is to calculate \(d u / d x\). The function given is \(u=5 x+9\). The derivative of \(x\) with respect to \(x\) is \(1\) and constant vanishes after being differentiated, so \(d u / d x = 5\).

Step 3: Derivative of y with respect to x

To find \(d y / d x = d y / d u * d u / d x\) by applying the chain rule. So substituting the values derived above, \(d y / d x = 1/\sqrt{u} * 5 = 5/\sqrt{u}\).\nRemember, \(u\) is actually \(u=5 x+9\), substituting \(u\) back into the derivative gives the final answer \(d y / d x = 5/\sqrt{5x+9}\).

Key Concepts

Understanding the Derivative

A derivative represents the rate at which a function is changing at any given point. It's like looking at the speed of a car at a specific moment. In the context of our exercise, what we're interested in is how rapidly the function changes. More specifically, we started by looking at the derivative of the function \( y = 2\sqrt{u} \) with respect to \( u \).

Let's break it down: the derivative of \( \sqrt{u} \) with respect to \( u \) is \( \frac{1}{2\sqrt{u}} \). This is because the power rule tells us how to handle expressions like \( u^{1/2} \). Remember, the power rule involves bringing down the exponent as a coefficient and then reducing the original exponent by one. Applying that rule here gives us \( \frac{1}{2}u^{-1/2} \), which translates to \( \frac{1}{2\sqrt{u}} \).

• For \( y = 2\sqrt{u} \), multiply \( 2 \) by \( \frac{1}{2\sqrt{u}} \) to yield \( \frac{1}{\sqrt{u}} \) as the result for \( \frac{dy}{du} \).

So, \( \frac{dy}{du} \) tells us how \( y \) changes with \( u \). It's all about rates of change.

Concept of Composite Functions

Composite functions involve applying one function to the result of another. Think of it as a series of steps where the output of one "machine" becomes the input of another.

In the original problem, we found ourselves dealing with \( y = 2\sqrt{u} \), where \( u = 5x + 9 \). Here, \( y \) is a function of \( u \), and \( u \) is a function of \( x \). Such nesting of functions is what gives us a composite function.

• Composite functions are essential because they allow the building of complex relationships from simpler ones.
• They necessitate applying the chain rule when finding derivatives, due to their layered nature.

This interconnected structure means that to understand the full story of how \( y \) changes with respect to \( x \), we need to consider both these layers of functions, i.e., \( y \) depending on \( u \) and \( u \) on \( x \). This brings us to the realm of differentiation through composite paths.

Differentiation and the Chain Rule

Differentiation is the process of finding the derivative, essentially uncovering the rate of change for functions. When functions are composite, differentiation requires a useful technique called the chain rule.

The chain rule is a formula for finding the derivative of a composite function. If you ever come across functions nested within each other, the chain rule is your go-to method. It's like peeling back layers of an onion, finding the derivative of each layer until you reach the innermost core.

In our exercise with \( y \) and \( u \), the chain rule lets us combine \( \frac{dy}{du} \) and \( \frac{du}{dx} \) to determine \( \frac{dy}{dx} \). The rule can be stated as: the derivative of the composite \( y(f(x)) \) is \( \frac{dy}{du} \times \frac{du}{dx} \). In simpler terms, it's the product of the derivative of the outer function evaluated at the inner function times the derivative of the inner function.

• For our case, \( \frac{dy}{dx} = \frac{1}{\sqrt{u}} \times 5 \).
• This carries through as \( \frac{5}{\sqrt{5x + 9}} \) once you substitute \( u = 5x + 9 \) back in.
• This answers how \( y \) finally changes relative to \( x \).

Through these steps, the chain rule simplifies and makes the differentiation of complex functions much more manageable.