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Chapter 3: Problem 2

Find the first and second derivatives. $$y=x^{2}+x+8$$

Short Answer

Expert verified
The first derivative is \(y' = 2x + 1\) and the second derivative is \(y'' = 2\).

Step by step solution

01

Differentiate to find the first derivative

To find the first derivative of the function, apply the power rule to each term. The power rule states that if you have a term ax^n, its derivative is anx^{n-1}. Apply this to each term: For the first term: \(x^2\) becomes \(2x^{2-1} = 2x\).For the second term: \(x\) becomes \(1x^{1-1} = 1\).The constant term \(8\) becomes \(0\) since the derivative of a constant is always zero.Thus, the first derivative \(y' = 2x + 1\).
02

Differentiate again to find the second derivative

Now, take the first derivative we found \(y' = 2x + 1\) and differentiate it to find the second derivative. Apply the power rule again:For the first term: \(2x\) becomes \(2x^{1-1} = 2\).The second term \(1\) is a constant, so its derivative is \(0\).Thus, the second derivative \(y'' = 2\).

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Key Concepts

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

First Derivative
In calculus, finding the first derivative of a function is crucial as it helps to determine the rate of change of the function with respect to its variable. For the function given in the exercise, which is \( y = x^2 + x + 8 \), we apply the power rule to each term to find the first derivative. The power rule is a simple and powerful tool in calculus that states: if you have a term in the form \( ax^n \), the derivative is \( anx^{n-1} \). In this case:
  • The term \( x^2 \) becomes \( 2x^{2-1} \) which simplifies to \( 2x \).
  • The term \( x \) can be seen as \( 1x^1 \), and using the power rule, it changes to \( 1x^{1-1} = 1 \).
  • The constant 8 has a derivative of 0, as constants do not change with respect to \( x \).
Putting it together, the first derivative of \( y \) is \( y' = 2x + 1 \). This expression \( y' \) tells us how \( y \) changes with \( x \).
Second Derivative
Once we have the first derivative, the next step is often to find the second derivative. The second derivative, denoted as \( y'' \), is simply the derivative of the first derivative. It provides information about the concavity and inflection points of the original function.Using our first derivative \( y' = 2x + 1 \), we apply the power rule again:
  • The term \( 2x \) simplifies to \( 2 x^{1-1} = 2 \).
  • The constant term \( 1 \) becomes 0 since the derivative of any constant is 0.
Therefore, the second derivative is \( y'' = 2 \). This result indicates that the original function's graph is a parabola opening upwards, as its second derivative is a positive constant, showing consistent concavity.
Power Rule
The power rule is one of the most fundamental rules in differentiation, making it easy to handle polynomial functions and simple terms.The power rule can be stated simply: if you have a power function \( ax^n \), the derivative is \( anx^{n-1} \). Here's why this is so helpful:
  • It allows for quick calculations without complex algebra.
  • It simplifies higher degree polynomials, making them manageable.
  • It's an essential foundation for calculus, leading into other rules like the product rule and chain rule.
In the given exercise, each term of the polynomial follows the power rule, transforming them from terms like \( x^2 \) and \( x \) into simpler expressions quickly, which aids in understanding how the function behaves.

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Most popular questions from this chapter

Graph \(y=-2 x \sin \left(x^{2}\right)\) for \(-2 \leq\) \(x \leq 3 .\) Then, on the same screen, graph $$y=\frac{\cos \left((x+h)^{2}\right)-\cos \left(x^{2}\right)}{h}$$ for \(h=1.0,0.7,\) and \(0.3 .\) Experiment with other values of \(h\) What do you see happening as \(h \rightarrow 0 ?\) Explain this behavior.

Find the domain and range of each composite Iunction. Then graph the composites on separate screens. Do the graphs make sense in each case? Give reasons for your answers. Comment on any differences you see. a. \(y=\tan ^{-1}(\tan x)\) b. \(y=\tan \left(\tan ^{-1} x\right)\)

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