Question

(a) Use the Product Rule to differentiate the function. (b) Manipulate the function algebraically and differentiate without the Product Rule. (c) Show that the answers from (a) and (b) are equivalent. $$f(x)=\left(x^{2}+5\right)\left(3-x^{3}\right)$$

Short answer

The derivative is \( -5x^4 - 15x^2 + 6x \), verified through both methods.

Step-by-step solution

Use Product Rule

The Product Rule states that if you have a function that is the product of two functions, say \( u(x) \) and \( v(x) \), then its derivative \( f'(x) \) is given by \( u'(x)v(x) + u(x)v'(x) \). For the function \( f(x) = (x^2 + 5)(3 - x^3) \), identify \( u(x) = x^2 + 5 \) and \( v(x) = 3 - x^3 \).\First, differentiate \( u(x) \) to get \( u'(x) = 2x \).\Next, differentiate \( v(x) \) to get \( v'(x) = -3x^2 \).\Now apply the Product Rule: \[ f'(x) = u'(x)v(x) + u(x)v'(x) = (2x)(3 - x^3) + (x^2 + 5)(-3x^2) \] Simplifying, we have: \[ f'(x) = 6x - 2x^4 - 3x^4 - 15x^2 \] Combine like terms: \[ f'(x) = -5x^4 - 15x^2 + 6x \].

Algebraic Manipulation

Let's multiply out the terms in the function \( f(x) = (x^2 + 5)(3 - x^3) \). Expand by distributing each term of the first polynomial over each term of the second polynomial:\[ f(x) = x^2(3) + x^2(-x^3) + 5(3) + 5(-x^3) \]This simplifies to:\[ f(x) = 3x^2 - x^5 + 15 - 5x^3 \].Now differentiate using the power rule, where the derivative of \( ax^n \) is \( nax^{n-1} \):\[ f'(x) = 6x - 5x^4 - 15x^2 \].

Compare and Verify

Now let's compare the derivatives obtained from Step 1 and Step 2. - From the Product Rule: \( f'(x) = -5x^4 - 15x^2 + 6x \).- From the algebraic manipulation: \( f'(x) = -5x^4 - 15x^2 + 6x \).Both expressions are identical, which confirms that the calculations were done correctly and the results are equivalent.

Key concepts

Derivative

When you are tasked with finding the rate at which a function changes, you calculate its derivative. In essence, the derivative provides a blueprint of how a function behaves when subject to change. For functions expressed as products of two separate functions, you'll often need to use special rules to find the derivative, like the Product Rule. This rule helps you determine the derivative when two functions are multiplied together.
What does the derivative tell us? Here’s a simple breakdown for better understanding:

• The derivative gives us the slope of the tangent at any point on a curve.
• It's a measure of how steep a curve is at a certain point.
• Provides insights into increasing or decreasing trends of the function.

For example, in the exercise above, you're asked to break down and analyze how to differentiate a function which is a product of two smaller functions, using detailed derivative rules.

Algebraic Manipulation

Algebraic manipulation involves changing the form of a function in a mathematical expression, but without altering its value. This technique is powerful for simplifying functions to make differentiation or integration more manageable.
In the given problem, the function is multiplied out before applying differentiation. This means you take each term from one bracket and multiply it by each term from another bracket. Doing this will break the product into individual terms, making it simpler to differentiate each term separately.
Here's a more detailed look into algebraic manipulation:

• Useful for simplifying complex expressions.
• Allows easy application of basic differentiation rules like the power rule.
• Helps in eliminating use of complex rules if the expression can be broken down.

For our function, distributing terms from a polynomial in this manner alleviated the complexity of the original expression.

Power Rule

The power rule is one of the cornerstone rules in calculus for finding derivatives. It states that for a function in the form of \( ax^n \), its derivative is \( nax^{n-1} \). In simpler terms, you multiply the coefficient \( a \) by the power \( n \), then decrease the original power by one.
In the exercise, once the function was algebraically manipulated and simplified, the power rule was perfectly suited to differentiate each term.

• The power rule simplifies derivatives of polynomials.
• It's straightforward: just multiply and reduce the exponent.
• It's handy for expressions broken down by algebraic manipulation.

When confronted with expressions such as \( 3x^2 \), \( -x^5 \) or \( 15 \), use the power rule to swiftly determine the derivative, which becomes a powerful, frequently-used tool in solving calculus problems.