Problem 39 - Brief Applied Calculus

Question

Use the Generalized Power Rule to find the derivative of each function. $$ g(z)=2 z\left(3 z^{2}-z+1\right)^{4} $$

Step-by-Step Solution

Answer

The derivative is \(g'(z) = 2(3z^2 - z + 1)^4 + 8z(3z^2 - z + 1)^3(6z - 1)\).

Step 1: Identify the Functions

The given function is a product of two parts: \(2z\) and \((3z^2 - z + 1)^4\). Here, \(u(z) = 2z\) and \(v(z) = (3z^2 - z + 1)^4\). We will use the product rule to differentiate the product of these two functions.

Step 2: Apply the Product Rule

The product rule states that the derivative \((f \cdot g)'\) is \(f' \cdot g + f \cdot g'\). Let \(u(z) = 2z\) and \(v(z) = (3z^2 - z + 1)^4\). Then, \((u \cdot v)' = u' \cdot v + u \cdot v'\).

Step 3: Differentiate \(u(z) = 2z\)

The derivative of \(u(z) = 2z\) with respect to \(z\) is \(u'(z) = 2\).

Step 4: Differentiate \(v(z) = (3z^2 - z + 1)^4\) using the Generalized Power Rule

The Generalized Power Rule states that if \(v(x) = (h(x))^n\), then \(v'(x) = n(h(x))^{n-1}h'(x)\). Here, \(h(z) = 3z^2 - z + 1\) and \(n = 4\). Find \(h'(z)\):\[h'(z) = \frac{d}{dz}(3z^2 - z + 1) = 6z - 1.\] Now apply the Generalized Power Rule: \[v'(z) = 4(3z^2 - z + 1)^3(6z - 1).\]

Step 5: Substitute the Derivatives into the Product Rule

Substitute \(u'(z)\) and \(v'(z)\) back into the product rule: \[(2z)' \cdot (3z^2 - z + 1)^4 + 2z \cdot 4(3z^2 - z + 1)^3(6z - 1).\]

Step 6: Simplify the Expression

Simplify the expression by carrying out the multiplication: \[2(3z^2 - z + 1)^4 + 8z(3z^2 - z + 1)^3(6z - 1).\] This gives the derivative of \(g(z)\).

Key Concepts

Derivative

A derivative in calculus refers to the rate at which a function changes as its input changes. Simply put, it calculates the slope or the rate of change of a curve at a particular point. When you're working with a derivative, you are essentially finding how one quantity shifts in response to the change in another quantity. Derivatives are represented using the notation \(f'(x)\) or \(\frac{df}{dx}\). For example, in our exercise, we are determining how the function \(g(z) = 2z(3z^2 - z + 1)^4\) changes as \(z\) varies.

Product Rule

The product rule is essential when differentiating products of two functions. It states: if you have a function \(h(x) = f(x) \cdot g(x)\), the derivative \(h'(x)\) is given by the formula \(f'(x)g(x) + f(x)g'(x)\). Essentially, this means:

• First, differentiate the first function and multiply by the second unchanged.
• Then, leave the first function as it is and multiply by the derivative of the second function.

In our example, we have two functions: \(u(z) = 2z\) and \(v(z) = (3z^2 - z + 1)^4\). By applying the product rule, we compute the derivative of their product.

Calculus Problem Solving

Solving calculus problems involves understanding the function behavior and applying various rules effectively. You must identify the type of function given or if it's a composite of different types. The approach is methodical:

• Recognize the patterns or rules applicable (like the product or power rules).
• Break down the function into smaller parts if it's complex.
• Apply differentiation rules step-by-step.

In this case, recognizing that \(g(z)\) is a product of two differentiable functions allows for structured application of the product rule, combined with the more specific power rule for the inner term.

Power Rule in Calculus

The power rule is straightforward but powerful in finding the derivative of expressions of the form \(x^n\). The basic power rule states that if you have \(f(x) = x^n\), the derivative is \(f'(x) = nx^{n-1}\). The generalized power rule extends this concept for functions like \((h(x))^n\), where \(f'(x) = n(h(x))^{n-1}h'(x)\). In our exercise, the term \((3z^2 - z + 1)^4\) is differentiated using the generalized power rule:

• Differentiate the exponent: multiply by \(n\).
• Decrease the power by one and keep the inner function as it is.
• Multiply by the derivative of the inner function \(h(x)\).

This method ensures that complex power expressions are systematically differentiated, as seen in the function \(v(z)\).