Question

Find the value of the derivative of the function at the given point. State which differentiation rule you used to find the derivative. $$ f(x)=x^{2}\left(3 x^{3}-1\right) $$

Short answer

The derivative of the function is \( 15x^{4} - 2x \), found through the use of the product rule of differentiation.

Step-by-step solution

Recognize the Differentiation Rule

The given function \(f(x)=x^{2}(3x^{3}-1)\) is a product of two functions of x, \(x^{2}\) and \(3x^{3}-1\). Therefore, the product rule of differentiation will be used. The product rule states that the derivative of the product of two functions is the derivative of the first multiplied by the second plus the first multiplied by the derivative of the second.

Differentiate Using the Product Rule

Applying the product rule, write the first function as \(u(x) = x^{2}\) and the second function as \(v(x) = 3x^{3}-1\). Find the derivatives \(u'(x)\) and \(v'(x)\), which are \(2x\) and \(9x^{2}\) respectively. Therefore, \[ f'(x) = u'(x)v(x) + u(x)v'(x) = 2x(3x^{3}-1) + x^{2}(9x^{2}) = 6x^{4} - 2x + 9x^{4} \] \[ f'(x) = (6 + 9)x^{4} - 2x = 15x^{4} - 2x\]

Evaluate the Derivative

To find the value of the derivative at a specific point, simply replace all instances of \(x\), in \(f'(x)\), with the \(x\)-coordinate of the given point. The exercise did not specify a point, but the derivative of the function, which can be evaluated at a specific \(x\), has been found.

Key concepts

Differentiation

Differentiation is a fundamental concept in calculus used to determine how a function changes at any point. Essentially, when you differentiate a function, you're finding its derivative, which gives you the rate of change or the slope of the function at any given point. This is especially useful in understanding how variables interact with one another.

The process of differentiation involves various rules, and one of the most important is the Product Rule. However, differentiation can involve simpler power rules and chain rules, depending on the complexity of the function. By differentiating, you can understand aspects like velocity, acceleration, and gradients, which are essential in fields ranging from physics to economics.

Understanding differentiation involves:

• Recognizing the function components.
• Identifying applicable rules.
• Calculating derivatives through basic formulas such as the Product Rule or Chain Rule.
• Interpreting the meaning of these derivatives in practical terms.

Derivative Evaluation

Once you've found the expression for a derivative, evaluating it involves substituting specific values to find the precise rate of change at those points. This is needed when you want to measure how fast something is changing at a particular instant or location.

In the example given, the derivative of the function was calculated: \[ f'(x) = 15x^{4} - 2x \]To evaluate this derivative at a specific point, say at \( x = a \), substitute \( a \) into the derivative equation. This will allow you to find the slope of the tangent line to the function at \( x = a \), helping with predictions in models or understanding trends in data.

Steps to perform derivative evaluation:

• Substitute the given point into the derivative equation.
• Simplify the result to get a numerical value.
• Interpret the significance in terms of the original problem, like speed or growth rate.

Algebraic Functions

Algebraic functions are composed of terms involving numbers, constants, variables, and arithmetic operations like addition, multiplication, and exponents. They often form the basis for much of calculus, as they are used to model relationships.The given function in the exercise, \( f(x) = x^2(3x^3 - 1) \), is an example of an algebraic function where multiplication and power decoratively combine variables in polynomial form. Understanding how these components interact is essential in differentiation.

The Product Rule, which is a cornerstone for finding the derivative of algebraic functions like this, involves:

• Identifying each part of the function—here, \( u(x) = x^2 \) and \( v(x) = 3x^3 - 1 \).
• Applying differentiation rules to each separate piece.
• Combining these derivatives using the Product Rule to obtain the total derivative.

Grasping these concepts is crucial for breaking down more complex equations and applying calculus concepts to real-world scenarios.