Question

Finding a Derivative In Exercises \(25-38\) , find the derivative of the algebraic function. $$ f(x)=\left(x^{3}-x\right)\left(x^{2}+2\right)\left(x^{2}+x-1\right) $$

Short answer

The derivative of the function \(f(x) = (x^3 - x)(x^2 + 2)(x^2 + x - 1)\) is obtainable by carefully applying the power rule and product rule, and simplifying the result.

Step-by-step solution

Identify the functions

Three algebraic functions to differentiate are \(f_1(x) = x^3 - x\), \(f_2(x) = x^2 + 2\), and \(f_3(x) = x^2 + x - 1\).

Utilize Power Rule

Apply the power rule to find the derivatives of \(f_1\), \(f_2\), and \(f_3\): \(f'_1(x) = 3x^2 - 1\), \(f'_2(x) = 2x\), and \(f'_3(x) = 2x + 1\).

Apply the Product Rule

Use the product rule for function \(f = f_1 \cdot f_2 \cdot f_3\), which requires applying the product rule twice. Firstly: \((f_1 f_2)' = f'_1 f_2 + f_1 f'_2\). Secondly: \(((f_1 f_2)f_3)' = (f_1 f_2)' f_3 + (f_1 f_2) f'_3\). Substitute the values using the derivates found in Step 2.

Simplify the Resulting Expression

After substituting the values, simplify the resulting expression by expanding and combining like terms.

Key concepts

Product Rule

Understanding how to find the derivative of a product of functions is essential in calculus, and this is where the Product Rule comes into play. The Product Rule is applied when you have a function that is the product of two other functions, say h(x) = f(x) \times g(x).

The rule states that the derivative of the product of two functions is the first function multiplied by the derivative of the second plus the second function multiplied by the derivative of the first. Mathematically, the Product Rule is expressed as follows:
\[ h'(x) = f(x)g'(x) + g(x)f'(x) \].
When applying the Product Rule to the given exercise, remember that you may need to apply it multiple times if there are more than two functions multiplied together, just as shown in the step-by-step solution with three functions.

Power Rule

When dealing with the differentiation of algebraic functions involving exponents, the Power Rule is a straightforward tool to use. It applies to functions of the form f(x) = x^n, where n is any real number.

The Power Rule says that the derivative of f(x) = x^n with respect to x is nx^{n-1}. This is written as \[ f'(x) = nx^{n-1} \].
In the context of our exercise, the Power Rule was applied to find the derivatives of each individual function before using the Product Rule. Each term's exponent is reduced by one, and that original exponent becomes a coefficient, simplifying the process of finding derivatives for algebraic functions significantly.

Algebraic Functions

An Algebraic Function is a type of function that can be expressed using polynomial expressions as well as operations like addition, subtraction, multiplication, division, and taking roots, such as square roots. Examples include linear functions like f(x) = 2x + 3, quadratic functions like g(x) = x^2 - 4x + 4, and higher-degree polynomial functions.

In this exercise, we encounter three algebraic functions that are multiplied together. To find the derivative of complex algebraic functions that are products of simpler functions, you'll often need both the Product Rule and the Power Rule. It is also important to practice simplifying these functions; combining like terms and expanding where necessary to get to the simplest form for taking derivatives.

Derivative of a Function

The Derivative of a Function represents the rate at which the function's value is changing at any given point. This concept is pivotal in calculus and helps us understand motion, growth, and a multitude of other changing phenomena.

In finding the derivative of a function, we apply different rules depending on the function's form, including the Product Rule and the Power Rule as shown in the provided exercise. The derivative expresses the slope of the tangent line to the function at any point, and it is essential for solving problems in fields like physics, economics, and engineering. Calculating the derivative is also the foundational step for more advanced concepts in calculus, such as integration and solving differential equations.