Problem 7 - Calculus Early Transcendentals

Question

Find \(f^{\prime}(x)\). \(f(x)=\left(x^{3}+7 x^{2}-8\right)\left(2 x^{-3}+x^{-4}\right)\)

Step-by-Step Solution

Answer

The derivative is \( f'(x) = -15x^{-2} - 14x^{-3} + 48x^{-4} + 32x^{-5} \).

Step 1: Identify the Function Types

First, observe the function given: \[ f(x) = (x^3 + 7x^2 - 8)(2x^{-3} + x^{-4}) \]. It is a product of two functions, so we will use the product rule to differentiate it.

Step 2: Recall the Product Rule

The product rule states that if \( u(x) \) and \( v(x) \) are functions of \( x \), then the derivative of their product \( u(x)v(x) \) is given by \( u'(x)v(x) + u(x)v'(x) \). Here, \( u(x) = x^3 + 7x^2 - 8 \) and \( v(x) = 2x^{-3} + x^{-4} \).

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

Compute the derivative of \( u(x) = x^3 + 7x^2 - 8 \):\[ u'(x) = 3x^2 + 14x \].

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

Compute the derivative of \( v(x) = 2x^{-3} + x^{-4} \): \[ v'(x) = -6x^{-4} - 4x^{-5} \].

Step 5: Apply the Product Rule

Apply the product rule: \[ f'(x) = u'(x)v(x) + u(x)v'(x) \]. Substitute the derivatives computed earlier: \[ f'(x) = (3x^2 + 14x)(2x^{-3} + x^{-4}) + (x^3 + 7x^2 - 8)(-6x^{-4} - 4x^{-5}) \].

Step 6: Simplify the Expression

Simplify the expression: 1. Calculate \( (3x^2 + 14x)(2x^{-3} + x^{-4}) = 3x^2\cdot2x^{-3} + 3x^2\cdot x^{-4} + 14x\cdot 2x^{-3} + 14x\cdot x^{-4} \). Simplifying each term: \( 6x^{-1} + 3x^{-2} + 28x^{-2} + 14x^{-3} \) which simplifies to \( 6x^{-1} + 31x^{-2} + 14x^{-3} \).2. Calculate \( (x^3 + 7x^2 - 8)(-6x^{-4} - 4x^{-5}) = x^3\cdot(-6x^{-4}) + x^3\cdot(-4x^{-5}) + 7x^2\cdot(-6x^{-4}) + 7x^2\cdot(-4x^{-5}) + (-8)\cdot(-6x^{-4}) + (-8)\cdot(-4x^{-5}) \). Simplifying each term: \( -6x^{-1} - 4x^{-2} - 42x^{-2} - 28x^{-3} + 48x^{-4} + 32x^{-5} \) which simplifies to \( -6x^{-1} - 46x^{-2} - 28x^{-3} + 48x^{-4} + 32x^{-5} \).3. Add the two results to obtain the final expression: \[ f'(x) = (6x^{-1} + 31x^{-2} + 14x^{-3}) + (-6x^{-1} - 46x^{-2} - 28x^{-3} + 48x^{-4} + 32x^{-5}) \]. Combine like terms to get: \[ f'(x) = 0x^{-1} - 15x^{-2} - 14x^{-3} + 48x^{-4} + 32x^{-5} \].

Key Concepts

Product Rule

In calculus differentiation, the product rule is a handy tool when dealing with the derivative of a product of two functions. Imagine this scenario: you have two functions, say \( u(x) \) and \( v(x) \), and you want to find the derivative of their product, \( u(x)v(x) \). The product rule provides a straightforward method to do this.

The rule states that the derivative of the product of \( u(x) \) and \( v(x) \) is \( u'(x)v(x) + u(x)v'(x) \), where \( u'(x) \) and \( v'(x) \) are the derivatives of \( u(x) \) and \( v(x) \) respectively.

• Differentiate \( u(x) \) to get \( u'(x) \).
• Differentiate \( v(x) \) to get \( v'(x) \).
• Apply the product rule formula \( u'(x)v(x) + u(x)v'(x) \).

The product rule is especially useful when each part of the function is complex enough to require individual derivatives. So, always remember, when you see a multiplication of functions in calculus, the product rule might be your best friend.

Derivatives of Polynomials

Polynomials are a fundamental part of calculus differentiation and finding their derivatives is essential. A polynomial is simply a mathematical expression involving a sum of powers of a variable with coefficients. Commonly, these powers are whole numbers.

When you're tasked with finding the derivative of a polynomial, you apply a straightforward rule: multiply the coefficient of each term by the exponent, then reduce the exponent by one. For example, the derivative of \( ax^n \) becomes \( nax^{n-1} \).

• Identify the highest power of \( x \) in each term.
• Apply the formula to each term of the polynomial.

Take the polynomial \( x^3 + 7x^2 - 8 \), for instance:

• The term \( x^3 \) becomes \( 3x^2 \).
• The term \( 7x^2 \) becomes \( 14x \).
• The constant \( -8 \) has a derivative of 0, as constants vanish when derived.

This method allows you to map out the changes in the slope of the polynomial function, giving you its rate of change at any point.

Negative Exponents

Negative exponents often come up in calculus, especially when dealing with power functions expressed in a more general form. Understanding them is crucial when differentiating such expressions. Simply put, a negative exponent means you take the reciprocal of the base raised to the opposite positive power. That is, \( x^{-n} = \frac{1}{x^n} \).

In the context of differentiation, treating negative exponents correctly is vital to finding derivatives accurately. After converting a negative exponent to a fraction, you differentiate using the power rule, just as you would with positive powers.

For instance, consider the function \( 2x^{-3} \). To differentiate, apply the power rule:

• Multiply the coefficient \( 2 \) by \(-3\), giving \(-6x^{-4}\).
• The new exponent is \(-4\) since you decrease the original exponent \(-3\) by one.

By handling negative exponents with confidence, you can simplify these expressions and proceed to differentiate them just like any other power term. Remember, embrace those negative exponents as opportunity to explore the inverse relationships within functions.