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Chapter 23: Problem 20

Find the derivative of each of the given functions. $$f(z)=-\frac{1}{4} z^{8}+\frac{1}{2} z^{4}-2^{3}$$

Short Answer

Expert verified
The derivative is \( f'(z) = -2z^7 + 2z^3 \).

Step by step solution

01

Identify the function

The given function is \( f(z) = -\frac{1}{4}z^8 + \frac{1}{2}z^4 - 2^3 \). We will find its derivative with respect to \( z \).
02

Simplify constants

Before differentiating, simplify any constants if possible. Here, \( 2^3 \) simplifies to 8. Thus, the function becomes \( f(z) = -\frac{1}{4}z^8 + \frac{1}{2}z^4 - 8 \).
03

Differentiate each term

Apply the power rule \( \frac{d}{dz}(z^n) = nz^{n-1} \) to each term:- For \( -\frac{1}{4}z^8 \), the derivative is \( -\frac{1}{4} \times 8z^{7} = -2z^{7} \).- For \( \frac{1}{2}z^4 \), the derivative is \( \frac{1}{2} \times 4z^{3} = 2z^{3} \).- For the constant \( -8 \), the derivative is 0.
04

Combine the derivatives

Combine all the derivatives from each term to find the total derivative:\( f'(z) = -2z^{7} + 2z^{3} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Power Rule
The power rule is a fundamental tool in calculus for finding the derivative of a polynomial function. It states that the derivative of a term in the form of \( z^n \), where \( n \) is a real number, is \( nz^{n-1} \). This rule is incredibly helpful because it simplifies the process of differentiation for many common functions.
  • For instance, applying the power rule to \( -\frac{1}{4}z^8 \) means multiplying the exponent 8 by the coefficient -1/4, resulting in \( -2z^7 \).
  • Similarly, \( \frac{1}{2}z^4 \) becomes \( 2z^3 \) after applying the power rule.
Learning to use the power rule effectively allows for quick calculations of derivatives, which is essential for solving more complex problems.
Function Differentiation
Differentiation is the process of finding the derivative of a function. The derivative measures how a function's output value changes as its input value changes. In simpler terms, it calculates the rate of change at any point of the function.
  • When differentiating, each term of the function is considered individually, as done in the original exercise. This technique ensures precision.
  • Constants in the function, such as -8 in the exercise, have derivatives of 0, as they do not change with the variable.
By systematically differentiating each part, you can find how complex functions evolve, which is critical for many applications in physics, engineering, and more.
Mathematical Expressions
Mathematical expressions are combinations of numbers, variables, operations, and sometimes symbols. A function like \( f(z) = -\frac{1}{4}z^8 + \frac{1}{2}z^4 - 8 \) is an example of a mathematical expression. Understanding how to manipulate these expressions is key to successfully applying calculus concepts such as differentiation.
  • Simplification, like turning \( 2^3 \) into 8, can make complex expressions more manageable.
  • Recognizing components of an expression, such as polynomials, constants, and variables, also guides which calculus tools to use.
By breaking down expressions into smaller parts, you can apply differentiation rules more clearly and accurately.
Calculus
Calculus is a branch of mathematics concerned with change and motion. It involves two main concepts: differentiation, which we have covered, and integration. Differentiation focuses on finding the slope of a function at any given point, reflecting the rate of change.
  • In our exercise, calculus is used to determine how the function \( f(z) \) changes as \( z \) changes.
  • Through calculus, we decipher various real-world phenomena, from the growth rate of populations to the way particles move in physics.
Mastering calculus opens up a deeper understanding of the world and numerous fields such as science and engineering.

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Most popular questions from this chapter

Solve the given problems by using implicit differentiation.The pressure \(P\), volume \(V\), and temperature \(T\) of a gas are related by \(P V=n(R T+a P-b P / T),\) where \(a, b, n,\) and \(R\) are \(c \subset\) stants. For constant \(V\), find \(d P / d T\).

Evaluate the indicated limits by evaluating the function for values shown in the table and observing the values that are obtained. Do not change the form of the function. $$\text { Find } \lim _{x \rightarrow \infty} \frac{1-x^{2}}{8 x^{2}+5}$$ $$\begin{array}{l|l|l|l} x & 10 & 100 & 1000 \\\\\hline f(x) & & &\end{array}$$

Evaluate the second derivative of the given function for the given value of \(x\). $$y=3 x^{2 / 3}-\frac{2}{x}, x=-8$$

Solve the given problems by finding the appropriate derivatives.A bullet is fired vertically upward. Its distance \(s\) (in \(\mathrm{ft}\) ) above the ground is given by \(s=2250 t-16.1 t^{2},\) where \(t\) is the time (in s). Find the acceleration of the bullet.

Solve the given problems by finding the appropriate derivatives.Show that $$\frac{d^{2}}{d x^{2}}(u v)=u \frac{d^{2} v}{d x^{2}}+2 \frac{d u}{d x} \frac{d v}{d x}+\frac{d^{2} u}{d x^{2}} v$$.
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