Chapter 2: Problem 40
Use the Generalized Power Rule to find the derivative of each function. $$ g(z)=z^{2}\left(2 z^{3}-z+5\right)^{4} $$
Short Answer
Expert verified
The derivative is \( g'(z) = 2z(2z^3 - z + 5)^4 + z^2 \cdot 4(2z^3 - z + 5)^3(6z^2 - 1) \).
Step by step solution
01
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 \( u'v + uv' \).
02
Differentiate the First Function \( u(z) = z^2 \)
Find \( u'(z) \):\( u'(z) = \frac{d}{dz}[z^2] = 2z \).
03
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:\( v'(z) = 4(2z^3 - z + 5)^{3} \cdot \frac{d}{dz}(2z^3 - z + 5) \).
04
Differentiate the Inner Function of \( v(z) \)
Find the derivative of the inner function \( 2z^3 - z + 5 \):\( \frac{d}{dz}(2z^3 - z + 5) = 6z^2 - 1 \).
05
Substitute Back to Find \( v'(z) \)
Substitute the derivative of the inner function back into \( v'(z) \):\( v'(z) = 4(2z^3 - z + 5)^{3}(6z^2 - 1) \).
06
Substitute \( u'(z) \) and \( v'(z) \) into the Product Rule
Substitute the derivatives back into the Product Rule formula:\( g'(z) = u'(z)v(z) + u(z)v'(z) \).
07
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:\[ g'(z) = 2z(2z^3 - z + 5)^4 + z^2 \cdot 4(2z^3 - z + 5)^{3}(6z^2 - 1). \]
08
Simplify
Simplify the expression to express \( g'(z) \) in a more concise form.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
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.
- The derivative of \( u(z)v(z) \) is \( u'(z)v(z) + u(z)v'(z) \).
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 \( z \).
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 \( 6z^2 - 1 \).
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 \( v'(z) \).