Problem 14 - University Calculus: Early Transcendentals

Question

In Exercises \(9-22,\) write the function in the form \(y=f(u)\) and \(u=g(x) .\) Then find \(d y / d x\) as a function of \(x\). $$y=\sqrt{3 x^{2}-4 x+6}$$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = \frac{6x - 4}{2\sqrt{3x^2 - 4x + 6}} \).

Step 1: Identify Inner and Outer Functions

Given the function \( y = \sqrt{3x^2 - 4x + 6} \), we need to identify the inner function \( u \) and the outer function \( y \). Define the inner function as \( u = g(x) = 3x^2 - 4x + 6 \). Define the outer function as \( y = f(u) = \sqrt{u} \).

Step 2: Compute the Derivative of the Outer Function

Now, compute the derivative of the outer function \( y = f(u) = \sqrt{u} \).The derivative of \( y \) with respect to \( u \) is given by \( \frac{dy}{du} = \frac{1}{2\sqrt{u}} \).

Step 3: Compute the Derivative of the Inner Function

Next, compute the derivative of the inner function \( u = 3x^2 - 4x + 6 \) with respect to \( x \).The derivative \( \frac{du}{dx} = 6x - 4 \).

Step 4: Use the Chain Rule to Find \( \frac{dy}{dx} \)

Apply the chain rule to find \( \frac{dy}{dx} \), which states that \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \).Substitute the derivatives computed:\[ \frac{dy}{dx} = \left( \frac{1}{2\sqrt{u}} \right) \times (6x - 4) \]Replace \( u \) back with \( 3x^2 - 4x + 6 \):\[ \frac{dy}{dx} = \frac{6x - 4}{2\sqrt{3x^2 - 4x + 6}} \]

Key Concepts

Derivative of a function

In calculus, the derivative of a function measures how a function's value changes as its input changes. It's essentially a way to compute the slope at any point on a curve. Understanding derivatives is vital for finding the rate at which something is changing. Think of it as the "instantaneous rate of change." For example, if you are driving a car and looking at the speedometer, the speed shown is essentially the derivative of your car's position concerning time.

The process of finding a derivative is called differentiation. Suppose we have a function like the one in the original exercise: \( y = \sqrt{3x^2 - 4x + 6} \). To find its derivative with respect to \( x \), we first identify the components (inner and outer functions) and then use differentiation rules, such as the chain rule, to break down complex expressions into simpler parts we can work with. The derivative, \( \frac{dy}{dx} \), gives us the rate at which \( y \) changes with a small change in \( x \).

Identifying the correct method to find the derivative is crucial, as it opens doors to understanding more advanced calculus topics. You don't just apply procedures mechanically; you truly grasp why they work.

Implicit Differentiation

Implicit differentiation is a powerful technique for finding derivatives of functions that are not easily solved for one variable in terms of another. In other words, it's useful when our function is not explicitly presented, such as \( y = f(x) \), but rather as an implicit relation between \( x \) and \( y \). This method allows us to differentiate both sides of an equation with respect to a variable simultaneously, typically \( x \).

For example, if the relationship between \( x \) and \( y \) is given by a more complex equation like \( x^2 + y^2 = 10 \), implicit differentiation helps to find \( \frac{dy}{dx} \) without first solving for \( y \). You differentiate both sides of the equation with respect to \( x \), naturally incorporating \( y \)'s derivative due to the chain rule:

• Differentiate each term separately and apply the chain rule where necessary (considering \( y \) as a function of \( x \)).
• Include \( \frac{dy}{dx} \) wherever a term of \( y \) is differentiated.
• Solve for \( \frac{dy}{dx} \) to discover how \( y \) changes as \( x \) does.

This method widens your toolbox when dealing with harder equations involving more than one variable.

Function Composition

Function composition is when you combine two functions to form a new one. If you have two functions, say \( f(x) \) and \( g(x) \), their composition \( f(g(x)) \) means you first apply \( g \) to \( x \), then take the result and apply \( f \) to it. It’s like plugging one entire function into another, analogous to an assembly line.

In the original exercise concerning \( y = \sqrt{3x^2 - 4x + 6} \), you identify the inner and outer functions:

• The inner function is \( u = g(x) = 3x^2 - 4x + 6 \).
• The outer function is \( y = f(u) = \sqrt{u} \).

To find the derivative of a composite function like this, you typically use the chain rule, which helps in differentiating complex functions by breaking them down into manageable pieces.

The beauty of function composition lies in its versatility. It's used not only in calculus but also in programming and systems engineering to simplify and solve complex operations. Recognizing when to employ function composition simplifies solving problems that might otherwise seem daunting.