Problem 40 - Brief Applied Calculus

Question

Use the Generalized Power Rule to find the derivative of each function. $$ g(z)=z^{2}\left(2 z^{3}-z+5\right)^{4} $$

Step-by-Step Solution

Answer

The derivative is \( g'(z) = 2z(2z^3 - z + 5)^4 + z^2 \cdot 4(2z^3 - z + 5)^3(6z^2 - 1) \).

Step 1: Apply the Product Rule

To differentiate the function \( g(z) = z^{2}(2z^{3} - z + 5)^{4} \), we first recognize it as a product of two functions: \( u(z) = z^2 \) and \( v(z) = (2z^3 - z + 5)^4 \). According to the Product Rule, the derivative of a product \( uv \) is \( u'v + uv' \).

Step 2: Differentiate the First Function \( u(z) = z^2 \)

Find \( u'(z) \):\( u'(z) = \frac{d}{dz}[z^2] = 2z \).

Step 3: Apply the Generalized Power Rule to \( v(z) = (2z^3 - z + 5)^4 \)

To find \( v'(z) \), we apply the Generalized Power Rule where we differentiate the outer function and multiply by the derivative of the inside function:\( v'(z) = 4(2z^3 - z + 5)^{3} \cdot \frac{d}{dz}(2z^3 - z + 5) \).

Step 4: Differentiate the Inner Function of \( v(z) \)

Find the derivative of the inner function \( 2z^3 - z + 5 \):\( \frac{d}{dz}(2z^3 - z + 5) = 6z^2 - 1 \).

Step 5: Substitute Back to Find \( v'(z) \)

Substitute the derivative of the inner function back into \( v'(z) \):\( v'(z) = 4(2z^3 - z + 5)^{3}(6z^2 - 1) \).

Step 6: Substitute \( u'(z) \) and \( v'(z) \) into the Product Rule

Substitute the derivatives back into the Product Rule formula:\( g'(z) = u'(z)v(z) + u(z)v'(z) \).

Step 7: Compute \( g'(z) \) Using the Product Rule Formula

Substitute \( u'(z) = 2z \), \( u(z) = z^2 \), \( v(z) = (2z^3 - z + 5)^4 \), and \( v'(z) = 4(2z^3 - z + 5)^{3}(6z^2 - 1) \) into the formula:\[ g'(z) = 2z(2z^3 - z + 5)^4 + z^2 \cdot 4(2z^3 - z + 5)^{3}(6z^2 - 1). \]

Step 8: Simplify

Simplify the expression to express \( g'(z) \) in a more concise form.

Key Concepts

Product Rule

The Product Rule is a crucial tool for finding the derivative of a product of two functions. In calculus, if you have two functions, say \( u(z) \) and \( v(z) \), their product, \( u(z)v(z) \), is a new function. The derivative of this product is not simply the product of their derivatives. Instead, the Product Rule states:

• The derivative of \( u(z)v(z) \) is \( u'(z)v(z) + u(z)v'(z) \).

Here, \( u'(z) \) and \( v'(z) \) represent the derivatives of \( u \) and \( v \) respectively.
This rule effortlessly handles the complexity that comes with multiplying two functions by breaking it into manageable parts.
It plays a vital role in the solution of our original problem, where we needed to find \( g'(z) \) by viewing \( g(z) = z^2(2z^3 - z + 5)^4 \) as a product of \( u(z) = z^2 \) and \( v(z) = (2z^3 - z + 5)^4 \). By substituting these into the Product Rule formula, we obtained a clear path to finding the derivative.

Derivative

Simply put, a derivative measures how a function changes as its input changes. It's a fundamental concept in calculus and provides insight into the rate at which things happen.
In mathematical terms, if you have a function \( f(x) \), its derivative, \( f'(x) \), gives you the slope of the tangent line to the graph of \( f \) at any point \( x \).
In the problem we tackled, finding the derivative was essential.
By identifying \( u(z) = z^2 \), we computed \( u'(z) = 2z \) using basic derivative rules.
Meanwhile, for \( v(z) = (2z^3 - z + 5)^4 \), the derivative involved both the derivative of the outer function via the Generalized Power Rule and the derivative of the inner function, resulting in a more complex calculation.
These derivatives were then combined using the Product Rule to discover how \( g(z) \) changes with respect to \( z \).

Inner Function

Often in calculus, you will deal with composite functions - functions within functions. The inner function refers to the part inside another function that needs attention when differentiating.
In our problem, the function \( v(z) = (2z^3 - z + 5)^4 \) is an example. Here, the expression \( 2z^3 - z + 5 \) is the inner function.
To differentiate such a nested function, we used the Generalized Power Rule, which involves taking care of both the outer function and the inner function separately.
After identifying the inner function \( 2z^3 - z + 5 \), its derivative was calculated as \( 6z^2 - 1 \).

• This derivative was then multiplied by the derivative of the outer function \( 4(2z^3 - z + 5)^3 \) to find the overall derivative \( v'(z) \).

Understanding the role of the inner function and accurately computing its effect on the overall derivative is crucial for solving such problems with confidence.