Problem 27 - Calculus Single Variable

Question

Compute \((f \circ g)^{\prime}\) and \((g \circ f)^{\prime}\). $$ f(x)=x^{4}+7 x^{2}, g(x)=\sqrt{x} $$

Step-by-Step Solution

Answer

\((f \circ g)'(x) = 2x + 7\) and \((g \circ f)'(x) = \frac{4x^3 + 14x}{2\sqrt{x^4 + 7x^2}}\)."

Step 1: Identify the Composition Functions

The task involves two composition functions: 1. \( (f \circ g)(x) = f(g(x)) = f(\sqrt{x}) \)2. \( (g \circ f)(x) = g(f(x)) = g(x^4 + 7x^2) \)

Step 2: Differentiate the Inner and Outer Functions

To differentiate these compositions using the chain rule, first find the derivatives:- The derivative of \( f(x) = x^4 + 7x^2 \) is \( f'(x) = 4x^3 + 14x \).- The derivative of \( g(x) = \sqrt{x} \) is \( g'(x) = \frac{1}{2\sqrt{x}} \).

Step 3: Apply the Chain Rule to (f ∘ g)′

Using the chain rule for the composition \( (f \circ g)(x) = f(g(x)) \):- Find the derivative \( (f \circ g)'(x) = f'(g(x)) \cdot g'(x) \).- Substitute the derivatives and functions: \( f'(\sqrt{x}) = 4(\sqrt{x})^3 + 14(\sqrt{x}) \). - Thus, \( (f \circ g)'(x) = (4x^{3/2} + 14x^{1/2}) \cdot \frac{1}{2\sqrt{x}} \).

Step 4: Simplify the Expression for (f ∘ g)′

Simplify the expression from Step 3:- \( (f \circ g)'(x) = \frac{4x^{3/2} + 14x^{1/2}}{2\sqrt{x}} \).- Simplify further: \( = 2x^1 + 7 \) or \( = 2x + 7 \) when divided across terms.

Step 5: Apply the Chain Rule to (g ∘ f)′

Now, apply the chain rule for \( (g \circ f)(x) = g(f(x)) \):- Differentiate: \( (g \circ f)'(x) = g'(f(x)) \cdot f'(x) \).- Calculate: \( g'(f(x)) = \frac{1}{2\sqrt{x^4 + 7x^2}} \).- So, \( (g \circ f)'(x) = \frac{1}{2\sqrt{x^4 + 7x^2}} \cdot (4x^3 + 14x) \).

Step 6: Simplify the Expression for (g ∘ f)′

Write in a more simplified form:- \( (g \circ f)'(x) = \frac{4x^3 + 14x}{2\sqrt{x^4 + 7x^2}} \).

Key Concepts

Derivative

The derivative is a fundamental concept in calculus and represents the rate at which a function changes at any given point. In simpler terms, the derivative tells us how a small change in the input of a function affects the output. When we talk about finding the derivative of a function, we are essentially looking for its slope or its rate of change.

• For a simple function like \( f(x) = x^4 + 7x^2 \), the derivative \( f'(x) \) is calculated by using rules like the power rule.
• The power rule states that the derivative of \( x^n \) is \( nx^{n-1} \). Thus, for \( x^4 \), the derivative is \( 4x^3 \).
• Similarly, for \( 7x^2 \), the derivative is \( 14x \).

Understanding derivatives is crucial because they form the building blocks for more advanced concepts such as the chain rule, which involves more complex compositions of functions.

Composition of Functions

When we deal with composition of functions, we are essentially plugging one function into another. This can seem confusing at first, but with practice, it becomes a very useful tool in calculus. Consider two functions \( f(x) \) and \( g(x) \). Their compositions are denoted as \( f(g(x)) \) and \( g(f(x)) \).

• For example, if \( f(x) = x^4 + 7x^2 \) and \( g(x) = \sqrt{x} \), then \( (f \circ g)(x) = f(g(x)) = f(\sqrt{x}) \) means you first apply \( g \) and then \( f \).
• Conversely, \( (g \circ f)(x) = g(f(x)) = \sqrt{x^4 + 7x^2} \) means you first apply \( f \) and then \( g \).

Compositions allow us to tackle more intricate functions by breaking them down into simpler ones that are easier to manage and analyze. This breakdown is particularly handy when differentiating complex expressions using the chain rule.

Differentiation Techniques

Differentiation techniques involve various rules and strategies to find the derivative of complex functions. The chain rule is one such essential technique. The chain rule is used to differentiate compositions of functions, which might not be straightforward otherwise. Here’s how it works in simple terms: If you have a composition \( (f \circ g)(x) = f(g(x)) \), the derivative is \( f'(g(x)) \cdot g'(x) \). This rule allows us to work inside out, differentiating the outer function first and multiplying it by the derivative of the inner function.

• For \( (f \circ g)'(x) = f'(g(x)) \cdot g'(x) \), we calculate \( f' \) at \( g(x) \) and multiply by \( g'(x) \) to get the final result.
• Similarly for \( (g \circ f)'(x) \), apply the rule by evaluating \( g' \) at \( f(x) \) and multiply by \( f'(x) \).

By thoroughly understanding and practicing these differentiation techniques, solving even the most complicated calculus problems becomes manageable and less daunting.