Problem 15 - The Calculus with Analytic Geometry

Question

Differentiate the given function by applying the theorems of this section. $$ f(x)=\left(2 x^{4}-1\right)\left(5 x^{3}+6 x\right) $$

Step-by-Step Solution

Answer

The derivative of \( f(x) \) is \( 70x^6 + 48x^4 - 3x^2 - 6 \).

Step 1: Identify the function

Given the function to differentiate is \[ f(x) = (2x^4 - 1)(5x^3 + 6x) \].This function is a product of two functions.

Step 2: Apply the product rule

To differentiate a product of two functions, use the product rule:\[ f'(x) = g'(x)h(x) + g(x)h'(x) \]where \[ g(x) = 2x^4 - 1 \] and \[ h(x) = 5x^3 + 6x \].

Step 3: Differentiate \(g(x)\)

Find the derivative of \[ g(x) = 2x^4 - 1 \].Using the power rule, \[ g'(x) = 8x^3 \].

Step 4: Differentiate \(h(x)\)

Find the derivative of \[ h(x) = 5x^3 + 6x \].Using the power rule, \[ h'(x) = 15x^2 + 6 \].

Step 5: Substitute into the product rule

Now substitute the derivatives into the product rule formula:\[ f'(x) = (8x^3)(5x^3 + 6x) + (2x^4 - 1)(15x^2 + 6) \].

Step 6: Expand and simplify

Expand the expression:\[ f'(x) = 40x^6 + 48x^4 + 30x^6 + 12x^2 - 15x^2 - 6 \].Combine like terms to simplify:\[ f'(x) = 70x^6 + 48x^4 - 3x^2 - 6 \].

Key Concepts

Product Rule

The Product Rule is essential in Calculus to differentiate functions that are the product of two or more simpler functions. When you have a function like \( f(x) = g(x) \cdot h(x) \), you use the product rule to find its derivative. The product rule formula is: \[ f'(x) = g'(x)h(x) + g(x)h'(x) \]. Here, \( g(x) \) and \( h(x) \) are the two functions being multiplied, and \( g'(x) \) and \( h'(x) \) are their respective derivatives.

Let's break it down:

• Identify your two functions within the product. In our exercise, \( g(x) = 2x^4 - 1 \) and \( h(x) = 5x^3 + 6x \).
• Differentiating \( g(x) \) gives \( g'(x) = 8x^3 \) using the power rule.
• Differentiating \( h(x) \) gives \( h'(x) = 15x^2 + 6 \) using the power rule.

After calculating these derivatives, substitute them back into the product rule formula: \[ f'(x) = g'(x)h(x) + g(x)h'(x) \ f'(x) = (8x^3)(5x^3 + 6x) + (2x^4 - 1)(15x^2 + 6) \]. By expanding and simplifying, you get the final derivative expression.

Power Rule

The Power Rule is a fundamental technique in Calculus for differentiating polynomial functions. The rule states: if you have \( f(x) = x^n \), then its derivative is \( f'(x) = nx^{n-1} \). This rule makes it straightforward to compute derivatives of functions where the variable is raised to any constant power.

For example, in our exercise:

• The derivative of \( 2x^4 \) is \( 8x^3 \) (since \( n = 4 \), \( 2 \cdot 4 \cdot x^{4-1} = 8x^3 \)).
• The derivative of \( 5x^3 \) is \( 15x^2 \) (since \( n = 3 \, 5 \cdot 3 \cdot x^{3-1} = 15x^2 \)).

Using this rule, you can easily find the derivatives needed for the product rule application. Just remember to differentiate each term and multiply the coefficients.

Calculus

Calculus is the branch of mathematics focusing on derivatives and integrals. It is integral to understanding and modeling change. Differentiation, a core concept in Calculus, involves finding the rate at which a function changes at any given point. This helps in analyzing functions in terms of slopes, velocities, and other rates of change.

In our exercise:

• We use differentiation to find how the function \( f(x) \) changes as \( x \) changes.
• The product rule and power rule are specific techniques within differentiation to handle more complex functions.

Once you're comfortable with these rules and concepts, you can tackle more advanced problems in Calculus.

Remember, practice is key. Work through many problems to understand these concepts deeply and start to see the patterns. With time, differentiating products of functions and other challenging tasks will become routine.