Problem 6 - Calculus Early Transcendentals

Question

Find \(f^{\prime}(x)\). \(f(x)=\left(2-x-3 x^{3}\right)\left(7+x^{5}\right)\)

Step-by-Step Solution

Answer

\(f'(x) = -7 - 63x^2 + 10x^4 - 6x^5 - 24x^7\)

Step 1: Identify the Derivative Formula

The function is a product of two functions: \(f(x) = u(x) \times v(x)\), where \(u(x) = 2 - x - 3x^3\) and \(v(x) = 7 + x^5\). We will use the product rule for derivatives: \((uv)' = u'v + uv'\).

Step 2: Differentiate \(u(x)\)

Calculate the derivative of \(u(x) = 2 - x - 3x^3\).\[u'(x) = 0 - 1 - 9x^2 = -1 - 9x^2\]

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

Calculate the derivative of \(v(x) = 7 + x^5\).\[v'(x) = 0 + 5x^4 = 5x^4\]

Step 4: Apply the Product Rule

Substitute \(u(x)\), \(v(x)\), \(u'(x)\), and \(v'(x)\) into the product rule formula:\[f'(x) = u'(x)v(x) + u(x)v'(x)\]\[f'(x) = (-1 - 9x^2)(7 + x^5) + (2 - x - 3x^3)(5x^4)\]

Step 5: Simplify the Expression

Expand and simplify:\((-1 - 9x^2)(7 + x^5) = -7 - x^5 - 63x^2 - 9x^7\)\((2 - x - 3x^3)(5x^4) = 10x^4 - 5x^5 - 15x^7\)Combine terms:\[f'(x) = -7 - x^5 - 63x^2 - 9x^7 + 10x^4 - 5x^5 - 15x^7\]Simplify to:\[f'(x) = -7 - 63x^2 + 10x^4 - 6x^5 - 24x^7\]

Step 6: Write the Final Derivative

The simplified expression for the derivative is:\[f'(x) = -7 - 63x^2 + 10x^4 - 6x^5 - 24x^7\]

Key Concepts

Understanding Derivatives

Derivatives are a fundamental part of calculus. They allow us to understand the rate at which things change. Through derivatives, we can explore how a function behaves at different points. For this exercise, we are asked to find the derivative of a function using the product rule.
The product rule is a special technique used when dealing with the derivative of a product of two functions. If we have a function defined as a product, say \(f(x) = u(x) \times v(x)\), the derivative \(f'(x)\) is given by the formula:

• \((uv)' = u'v + uv'\)

Here, \(u'\) and \(v'\) represent the derivatives of \(u(x)\) and \(v(x)\), respectively. By applying this formula, we can determine how the product of these two functions changes with respect to \(x\).
In our example, we identified \(u(x) = 2 - x - 3x^3\) and \(v(x) = 7 + x^5\). Then, we found their derivatives: \(u'(x) = -1 - 9x^2\) and \(v'(x) = 5x^4\). This calculation sets the foundation for applying the product rule accurately.

Exploring Polynomial Functions

Polynomial functions are expressions that involve terms raised to various powers, typically having the form \(a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0\). They are prevalent in calculus because of their straightforward nature and easy differentiability.
In our exercise, both functions \(u(x)\) and \(v(x)\) are polynomials. Understanding the basic structure of a polynomial helps to easily compute its derivative. For \(u(x) = 2 - x - 3x^3\), each term can be tackled individually.

• The constant \(2\) has a derivative of zero.
• The term \(-x\) gives a derivative of \(-1\).
• The term \(-3x^3\) becomes \(-9x^2\) when derived.

Similarly, for \(v(x) = 7 + x^5\), the derivative of the constant \(7\) is zero, while the derivative of \(x^5\) is \(5x^4\). Recognizing these patterns in polynomial differentiation can simplify complex expressions.

Mastering Calculus Techniques with Product Rule

Calculus techniques like using the product rule enable us to tackle complex functions, calculating their derivatives efficiently.
To apply the product rule, we carefully follow a structured approach. First, identify the functions involved in the product. Here, we have \(u(x)\) and \(v(x)\). Next, find each function's derivative, \(u'(x)\) and \(v'(x)\). This involves applying basic derivative rules, like for polynomials.
Once we have all derivatives, substitute into the product rule formula:

• \(f'(x) = u'(x)v(x) + u(x)v'(x)\)

For our function, this means computing

• \((-1 - 9x^2)(7 + x^5)\) and
• \((2 - x - 3x^3)(5x^4)\)

Finally, simplify the expression to reach the precise form of the derivative. This simplification helps in understanding how quickly the product function changes at any given point along \(x\).
These steps ensure that you are applying calculus techniques effectively to derive functions that might initially appear challenging.