Problem 7 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

Find the derivative of the given function. $$ f(x)=\left(5 x^{4}-3 x^{2}+2 x+1\right)^{10} $$

Step-by-Step Solution

Answer

The derivative is \[f'(x) = 10(5x^4 - 3x^2 + 2x + 1)^9 \times (20x^3 - 6x + 2).\]

Step 1: Identify the Function and Outer Function

The given function is a composite function where the outer function is raised to the 10th power. Let the inner function be denoted by \[u(x) = 5x^4 - 3x^2 + 2x + 1\] and the outer function be \[g(x) = (u(x))^{10}.\]

Step 2: Apply the Chain Rule

To differentiate the composite function \[f(x) = (u(x))^{10},\] use the chain rule. According to the chain rule:\[f'(x) = g'(u(x)) \times u'(x).\]

Step 3: Differentiate the Outer Function

Differentiate the outer function \[g(x) = (u(x))^{10}.\] The derivative of \[(u(x))^{10}\] with respect to \[u(x)\] is:\[g'(u) = 10(u(x))^{9}.\]

Step 4: Differentiate the Inner Function

Now, differentiate the inner function \[u(x) = 5x^4 - 3x^2 + 2x + 1.\] The derivative is calculated as:\[u'(x) = 20x^3 - 6x + 2.\]

Step 5: Combine the Results

Substitute \[u(x)\] and \[u'(x)\] into the chain rule formula:\[f'(x) = 10(u(x))^9 \times u'(x).\] Therefore:\[f'(x) = 10(5x^4 - 3x^2 + 2x + 1)^9 \times (20x^3 - 6x + 2).\]

Key Concepts

The Chain Rule

In calculus, the Chain Rule is a fundamental technique used to differentiate composite functions. Composite functions are functions within other functions. Consider the function $$f(x)=\big(5x^4-3x^2+2x+1\big)^{10}.$$ Here, the inner function is $$u(x)=5x^4-3x^2+2x+1$$ and the outer function is $$g(u)=u^{10}.$$ The Chain Rule states that to find the derivative of a composite function, you multiply the derivative of the outer function by the derivative of the inner function. Hence, $$f'(x)=g'(u(x))\times u'(x).$$ Applying the Chain Rule to our function, we get $$f'(x)=10\big(5x^4-3x^2+2x+1\big)^9\times (20x^3-6x+2).$$ The Chain Rule simplifies the process of differentiation for complex functions. By breaking the functions down and separately differentiating the inner and outer parts, we make challenging problems easier to handle.

Differentiation

Differentiation is a core concept in calculus that deals with finding the rate at which a function is changing at any given point. For simple functions, the process is straightforward, but it becomes more complex for composite functions. For example, when we want to differentiate $$f(x)=\big(5x^4-3x^2+2x+1\big)^{10},$$ we need to use both the Chain Rule and basic differentiation techniques. First, we identify the functions involved. The inner function is $$u(x)=5x^4-3x^2+2x+1,$$ and the outer function is $$g(u)=u^{10}.$$ Next, we find the derivatives of these functions separately. For the outer function, $$g'(u)=10u^9.$$ For the inner function, $$u'(x)=20x^3-6x+2.$$ Finally, we apply the Chain Rule: $$f'(x)=10\big(5x^4-3x^2+2x+1\big)^9\times(20x^3-6x+2).$$ Differentiation allows us to understand the behavior of functions, especially how they change and respond to variations in their variables. It's a vital tool in fields such as physics, engineering, economics, and beyond.

Understanding Calculus

Calculus is the branch of mathematics that studies how things change. It is divided into two main areas: differentiation and integration. Differentiation focuses on finding the rate of change—like how fast something is speeding up or slowing down. Integration deals with finding the total amount when the rate of change is known. When dealing with complex functions, calculus allows us to break down and analyze each component. For instance, to differentiate $$f(x)=\big(5x^4-3x^2+2x+1\big)^{10},$$ we use the Chain Rule to manage the layers of functions. This is crucial for solving real-world problems where variables are interconnected. Some key points to understand about calculus include:

• Limits: Understanding the value that a function approaches as the input approaches some value. This concept is foundational to defining derivatives.
• Continuity: A function is continuous if there are no breaks, jumps, or holes in its graph. This property is important for differentiability.
• Derivatives: These represent the slope or rate of change of a function at any point.

Calculus is a powerful tool for modeling and solving many types of problems in science, engineering, and economics. By mastering its fundamental concepts, you can unlock a deeper understanding of the world around you.