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Chapter 23: Problem 2

Make the given changes in the indicated examples of this section and then find the derivatives. In Example \(4,\) change \(3-x^{3}\) to \(2+x^{5}.\)

Short Answer

Expert verified
The derivative of the modified function is \( 5x^4 \).

Step by step solution

01

Identify the Original Function

In Example 4, we have an original function that is similar to something like \( y = 3 - x^3 \). Our task is to modify this to \( y = 2 + x^5 \). We need to focus on changing the expression within the original format.
02

Apply the Change

Change the given expression from \( 3 - x^3 \) to \( 2 + x^5 \). The modified function we need to differentiate is now \( y = 2 + x^5 \).
03

Apply the Derivative Rule

To find the derivative of the function \( y = 2 + x^5 \), we apply the power rule. The power rule states that the derivative of \( x^n \) is \( nx^{n-1} \). Since the derivative of a constant is zero, we focus only on \( x^5 \).
04

Differentiate Each Term

Calculate the derivative of \( x^5 \) with respect to \( x \):- The derivative of \( 2 \) is \( 0 \).- The derivative of \( x^5 \) is \( 5x^4 \).Therefore, the derivative \( y' = 0 + 5x^4 = 5x^4 \).
05

Final Derivative Function

After differentiating both terms, we find that the final derivative of the function \( y = 2 + x^5 \) is \( y' = 5x^4 \). This is the slope of the tangent line at any point on the curve for this function.

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Key Concepts

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

Understanding Derivatives
A derivative is a fundamental concept in calculus, representing the rate at which a function changes at any given point. Think of it as the "slope" of a function at a specific point on its curve. This slope indicates how steep the curve is. For instance, if you're climbing a hill, the derivative tells you how steep the hill is at each point.
  • The derivative notation is typically written as \( f'(x) \) or \( \frac{dy}{dx} \).
  • It helps in understanding how functions behaves and react to small changes.
  • In real-world terms, derivatives can represent speed, growth rates, and decline rates.
Grasping derivatives equips you to predict and comprehend myriad processes, from physics to economics.
Power Rule Simplifies Differentiation
The power rule is a quick technique to find the derivative of polynomial functions, like terms of the form \( x^n \). It's straightforward to use:
  • If \( f(x) = x^n \), then the derivative is \( f'(x) = nx^{n-1} \).
  • For example, with \( x^5 \), bringing down the 5 in front and reducing the power by one gives \( 5x^4 \).
This rule allows you to differentiate polynomial functions without complicated calculations. It's one of the most used shortcuts in calculus, saving time and effort when dealing with polynomial expressions.
Basics of Differentiation Techniques
Differentiation is the mathematical process of finding a derivative. It involves applying rules systematically to find the rate of change:
  • Start by identifying each term within the function that needs differentiation.
  • Apply differentiation rules like the power rule to each eligible term.
  • Simplify the result to get the derivative of the entire function.
Differentiation lets us determine how functions behave under various conditions, helping in modeling and predicting real-world scenarios. It forms the backbone of calculus, supporting advanced topics such as integration and differential equations.
Function Modification in Calculus
Function modification involves altering a function to explore new scenarios or solve specific problems. In our case, modifying \( 3-x^3 \) to \( 2+x^5 \) gives a new perspective on how these functions behave:
  • It often changes the behavior or properties of the original function.
  • Any modification requires recalculating derivatives and other properties.
  • By exploring modified functions, you gain insights into various aspects of calculus.
Changing functions and analyzing their derivatives is a common practice in calculus. It aids in understanding impacts of different variables or constants, crucial for mathematical modeling and problem-solving.

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Most popular questions from this chapter

Evaluate the function at \(0.1,0.01,\) and 0.001 from both sides of the value it approaches. In Exercises 53 and 54 evaluate the function for values of \(x\) of \(10,100,\) and \(1000 .\) From these values, determine the limit. Then, by using an appropriate change of algebraic form, evaluate the limit directly and compare values. $$\lim _{x \rightarrow 3} \frac{2 x^{2}-6 x}{x-3}$$

Evaluate the indicated limits by direct evaluation as in Examples \(10-14 .\) Change the form of the function where necessary. $$\lim _{x \rightarrow 4} \sqrt{x^{2}-7}$$

Solve the given problems by finding the appropriate derivatives.The voltage \(V\) induced in an inductor in an electric circuit is given by \(V=L\left(d^{2} q / d t^{2}\right),\) where \(L\) is the inductance (in \(\mathrm{H}\) ). Find the expression for the voltage induced in a \(1.60-\mathrm{H}\) inductor if \(q=\sqrt{2 t+1}-1\).

Solve the given problems by using implicit differentiation. A computer is programmed to draw the graph of the implicit function \(\left(x^{2}+y^{2}\right)^{3}=64 x^{2} y^{2}\) (see Fig. 23.45 and Example 7 on page 607 ). Find the slope of a line tangent to this curve at (2.00,0.56) and at (2.00,3.07)

Solve the given problems involving limits. Velocity can be found by dividing the displacement \(s\) of an object by the elapsed time \(t\) in moving through the displacement. In a certain experiment, the following values were measured for the displacements and elapsed times for the motion of an object. Determine the limiting value of the velocity. $$\begin{array}{l|c|c|c|c|c}s(\mathrm{cm}) & 0.480000 & 0.280000 & 0.029800 & 0.0029980 & 0.00029998 \\ \hline t(\mathrm{s}) & 0.200000 & 0.100000 & 0.010000 & 0.0010000 & 0.00010000\end{array}$$
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