Problem 1 - Thomas Calculus

Question

In Exercises \(1-8,\) given \(y=f(u)\) and \(u=g(x),\) find \(d y / d x=\) \(d y / d x=f^{\prime}(g(x)) g^{\prime}(x)\) $$ y=6 u-9, \quad u=(1 / 2) x^{4} $$

Step-by-Step Solution

Answer

The derivative \( \frac{dy}{dx} = 12x^3 \).

Step 1: Identify Functions

The problem gives us two functions: \( y = f(u) \) where \( y = 6u - 9 \), and \( u = g(x) \) where \( u = \frac{1}{2}x^4 \). Our goal is to find \( \frac{dy}{dx} \).

Step 2: Differentiate Outer Function

Differentiate the outer function \( y = 6u - 9 \) with respect to \( u \). This gives us \( \frac{dy}{du} = 6 \) since the derivative of a constant (\(-9\)) is zero.

Step 3: Differentiate Inner Function

Now, differentiate the inner function \( u = \frac{1}{2}x^4 \) with respect to \( x \). Using the power rule, we get \( \frac{du}{dx} = 2x^3 \).

Step 4: Apply Chain Rule

Use the chain rule, which states \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \). Substitute \( \frac{dy}{du} = 6 \) and \( \frac{du}{dx} = 2x^3 \) into the equation: \[ \frac{dy}{dx} = 6 \times 2x^3 = 12x^3 \]

Step 5: Simplify the Derivative

No further simplification is necessary. The derivative \( \frac{dy}{dx} \) is already in its simplest form.

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus, used to find the rate at which a function is changing at any point. When dealing with problems in calculus, you will often encounter the need to differentiate. Differentiation involves computing the derivative of a function and is critical when analyzing how one variable changes concerning another.

For example, if you have a function representing your position over time, the derivative of this function represents your speed. Differentiation allows us to:

• Find the slope of a tangent line to the graph of a function.
• Identify rates of change of quantities.
• Solve problems related to motion and growth.

The key rules for differentiation include the power rule, product rule, quotient rule, and the chain rule, each serving a specific type of problem. In exercises, you often utilize these rules, such as applying the power rule for polynomial functions and the chain rule for composite functions.

Composite Functions

Composite functions are essentially functions within functions. These are critical when dealing with complex problems that require multiple stages of computation. A composite function connects one function to the result of another, typically written as \( f(g(x)) \).

In the original exercise, we have two functions: \( y = 6u - 9 \) and \( u = \frac{1}{2}x^4 \). Together, they form a composite function where \( u = g(x) \) is substituted into \( y = f(u) \). This action leads us to analyze how a change in \( x \) affects the function \( y \) through the intermediary \( u \).

When dealing with composite functions, the chain rule is pivotal in calculating derivatives. This rule helps differentiate the "outer" function with respect to the "inner" function and then differentiates the "inner" function concerning \( x \). By understanding how composite functions build complexity from simpler components, you improve your problem-solving flexibility in calculus.

Power Rule

The power rule is a simple yet powerful tool for differentiation, especially when dealing with polynomials. It states that if you have a function \( u(x) = x^n \), its derivative is computed as \( \frac{du}{dx} = nx^{n-1} \).

In our example, the inner function \( u = \frac{1}{2}x^4 \) is a polynomial function, and by using the power rule, we can swiftly find its derivative: \( \frac{du}{dx} = 2x^3 \). Here's how it works:

• Identify the exponent (\( n = 4 \) in this case).
• Multiply by the constant coefficient (\( \frac{1}{2} \)), modifying the approach to \( 4 \times \frac{1}{2} \).
• Reduce the exponent by one to obtain the final expression \( 2x^3 \).

Utilizing the power rule effectively simplifies the process of finding derivatives, especially as polynomials are a common component in calculus problems. It complements other differentiation techniques, such as the chain rule used here, enhancing the overall problem-solving approach.