Problem 3 - Calculus - AP Edition

Question

Let \(f(x)=x^{5}\) and \(g(x)=2 x-3\) (a) Find \((f \circ g)(x)\) and \((f \circ g)^{\prime}(x)\) (b) Find \((g \circ f)(x)\) and \((g \circ f)^{\prime}(x)\)

Step-by-Step Solution

Answer

\((f \circ g)(x) = (2x - 3)^5\), \((f \circ g)'(x) = 10(2x - 3)^4\); \((g \circ f)(x) = 2x^5 - 3\), \((g \circ f)'(x) = 10x^4\).

Step 1: Understand Composition of Functions

The composition of functions \((f \circ g)(x)\) is defined as \(f(g(x))\). This means you substitute \(g(x)\) into every \(x\) in the function \(f(x)\). Similarly, \((g \circ f)(x)\) means substitute \(f(x)\) into every \(x\) in \(g(x)\).

Step 2: Find (f ∘ g)(x)

Let's substitute \(g(x)\) into \(f(x)\):\[f(g(x)) = ((2x - 3))^5\]. So, \((f \circ g)(x) = (2x - 3)^5\).

Step 3: Differentiate (f ∘ g)(x)

To find \((f \circ g)'(x)\), use the chain rule. Let \(u = 2x - 3\), so \(f(g(x)) = u^5\). Then the derivative \( (f \circ g)'(x) = \frac{d}{dx}(u)^5 = 5u^4 \cdot \frac{du}{dx}\). Since \(\frac{du}{dx} = 2\), we get \((f \circ g)'(x) = 5(2x - 3)^4 \cdot 2 = 10(2x - 3)^4\).

Step 4: Find (g ∘ f)(x)

Now substitute \(f(x)\) into \(g(x)\):\[g(f(x)) = 2(x^5) - 3\]. So, \((g \circ f)(x) = 2x^5 - 3\).

Step 5: Differentiate (g ∘ f)(x)

To find \((g \circ f)'(x)\), differentiate \(2x^5 - 3\) with respect to \(x\). By applying the power rule, the derivative of \(x^n\) is \(nx^{n-1}\). Therefore, \((g \circ f)'(x) = \frac{d}{dx}(2x^5 - 3) = 10x^4\).

Key Concepts

Chain Rule

The Chain Rule is an essential technique in calculus for finding the derivative of composite functions. A composite function is when one function is nested inside another, such as \((f \circ g)(x) = f(g(x))\). To effectively differentiate this type of function, the Chain Rule provides a systematic approach by breaking the problem into parts.Here's how it works:

• First, identify the "outer" function and the "inner" function. For example, in the function \(f(g(x))\), \(f\) is the outer function and \(g(x)\) is the inner function.
• Take the derivative of the outer function, treating the inner function \(g(x)\) as a constant. This derivative is often expressed as \(f'(g(x))\).
• Multiply this result by the derivative of the inner function, \(g'(x)\). The final derivative of the composite function is \((f \circ g)'(x) = f'(g(x)) \cdot g'(x)\).

For the exercise \((f \circ g)(x) = (2x - 3)^5\), applying the Chain Rule means first differentiating \((x)^5\), then multiplying by the derivative of \(2x - 3\), which is \(2\). This results in: \(10(2x - 3)^4\).
This systematic approach allows for efficient computation of derivatives in nested functions.

Derivative of Polynomials

Polynomial derivatives are a fundamental concept in calculus. They involve determining the rate at which a polynomial function changes. Calculating derivatives of polynomial functions is typically straightforward, thanks to some foundational rules.Let's break it down:

• A polynomial is a simple mathematical expression made up of variables, exponents, constants, and coefficients, such as \(a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0\).
• The derivative of a polynomial function finds the rate of change, or slope, at any given point along the polynomial's curve.
• In our exercise, the composite \((g \circ f)(x) = 2x^5 - 3\), involves finding the derivative by applying the rule to each term separately. The derivative of \(2x^5\) is \(10x^4\) (after applying the Power Rule), and the constant \(-3\) has a derivative of \(0\).
• The overall derivative is then simply the sum of these results.

This process makes it easy to work with polynomial equations and understand how they behave over intervals.

Power Rule

The Power Rule is one of the most important tools in calculus for finding derivatives. It simplifies the derivative-finding process by providing a direct formula for any power of \(x\).Here's the idea:

• The Power Rule states that for any term \(ax^n\), the derivative \(\frac{d}{dx}(ax^n) = nax^{n-1}\).
• This means you bring down the exponent \(n\) as a multiplier, then subtract one from the exponent to adjust the new power of \(x\).
• For example, in \(f(x) = x^5\), applying the Power Rule gives \(5x^4\) as the derivative because 5 (the exponent) is multiplied by \(x\) raised to the 4th power (exponent minus one).
• In the exercise's function \((g \circ f)(x) = 2x^5 - 3\), applying the Power Rule leads to derivatives for each term: \(2x^5\) becomes \(10x^4\), and \(-3\), being constant, becomes \(0\).

This simple rule is part of why calculus can be so incredibly powerful, making it possible to quickly handle complex polynomials and understand their rates of change.