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Chapter 3: Problem 10

Find the first and second derivatives. $$y=4-2 x-x^{-3}$$

Short Answer

Expert verified
First derivative: \(y' = -2 + 3x^{-4}\); Second derivative: \(y'' = -12x^{-5}\).

Step by step solution

01

Differentiate term by term

To find the first derivative of the function, differentiate each term of \(y = 4 - 2x - x^{-3}\) separately. The derivative of a constant, like \(4\), is \(0\). The derivative of \(-2x\) is \(-2\), and the derivative of \(-x^{-3}\) can be found using the power rule, which states the derivative of \(ax^n\) is \(anx^{n-1}\).
02

Apply the power rule to \(-x^{-3}\)

Use the power rule on the term \(-x^{-3}\). Here, \(a = -1\) and \(n = -3\), so the derivative is \(-1(-3)x^{-3-1} = 3x^{-4}\).
03

Summarize the first derivative

Combine the derivatives from each term to write the full first derivative: \(y' = 0 - 2 + 3x^{-4}\). Simplify to get \(y' = -2 + 3x^{-4}\).
04

Differentiate the first derivative term by term

Now, differentiate the first derivative \(y' = -2 + 3x^{-4}\) to find the second derivative. The derivative of \(-2\) is \(0\) since it is a constant, while \(3x^{-4}\) can be differentiated using the power rule.
05

Apply the power rule to \(3x^{-4}\)

For \(3x^{-4}\), apply the power rule: \(3(-4)x^{-4-1} = -12x^{-5}\).
06

Summarize the second derivative

Combine the results to write the full second derivative: \(y'' = 0 - 12x^{-5}\). Simplifying gives \(y'' = -12x^{-5}\).

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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 derivatives. It offers a quick method to differentiate functions of the form \(ax^n\), where \(a\) is a constant and \(n\) is a real number.
Essentially, when you have a term like \(x^n\), to find its derivative, you multiply \(n\) by the coefficient \(a\) and reduce the exponent by 1. In formula terms:
  • The derivative \((ax^n)' = anx^{n-1}\).
This rule simplifies the differentiation process significantly.
For example, to differentiate \(-x^{-3}\), you use \(a = -1\) and \(n = -3\). Apply the power rule:
  • Multiply \(-1\) by \(-3\), giving \(3\),
  • Subtract 1 from \(-3\), resulting in \(-4\).
The derivative simplifies to \(3x^{-4}\).
It鈥檚 important because it provides a straightforward, formulaic approach to differentiating power functions.
First Derivative
The first derivative of a function provides the rate at which the function's value is changing. It tells us how the function \(y\) is changing with respect to \(x\).
To find the first derivative of \(y = 4 - 2x - x^{-3}\), you differentiate each term separately:
  • Constant \(4\) becomes \(0\) because it does not change.
  • The term \(-2x\) becomes \(-2\).
  • Using the power rule, \(-x^{-3}\) becomes \(3x^{-4}\).
Combine these results to get the first derivative: \[ y' = 0 - 2 + 3x^{-4} \] Which simplifies to \(y' = -2 + 3x^{-4}\).
Studying the first derivative helps us understand the slope and direction of the function. Is it increasing or decreasing?
Second Derivative
The second derivative gives insight into the curvature of the original function \(y\). It tells you how the rate of change itself is changing.
If you already have the first derivative \(y' = -2 + 3x^{-4}\), you take the derivative again:
  • A constant rate like \(-2\) becomes \(0\).
  • Use the power rule to differentiate \(3x^{-4}\), transforming it into \(-12x^{-5}\).
The final second derivative is: \[ y'' = 0 - 12x^{-5} \] Which simplifies to \(y'' = -12x^{-5}\).
Understanding the second derivative helps determine if the function's graph is concave up or down, providing insight into the function's overall behavior.

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Most popular questions from this chapter

Suppose that the functions \(f\) and \(g\) and their derivatives with respect to \(x\) have the following values at \(x=0\) and \(x=1\) $$\begin{array}{ccccc} \hline x & f(x) & g(x) & f^{\prime}(x) & g^{\prime}(x) \\ \hline 0 & 1 & 1 & 5 & 1 / 3 \\ 1 & 3 & -4 & -1 / 3 & -8 / 3 \\ \hline \end{array}$$ Find the derivatives with respect to \(x\) of the following combinations at the given value of \(x\) a. \(5 f(x)-g(x), \quad x=1\) b. \(f(x) g^{3}(x), \quad x=0\) c. \(\frac{f(x)}{g(x)+1}, \quad x=1\) d. \(f(g(x)), \quad x=0\) e. \(g(f(x)), \quad x=0\) f. \(\left(x^{11}+f(x)\right)^{-2}, \quad x=1\) g. \(f(x+g(x)), \quad x=0\)

Find \(d y / d t\) $$y=\sin ^{2}(\pi t-2)$$

Find the value of \((f \circ g)^{\prime}\) at the given value of \(x\). $$f(u)=\left(\frac{u-1}{u+1}\right)^{2}, \quad u=g(x)=\frac{1}{x^{2}}-1, \quad x=-1$$

Give the position function \(s=f(t)\) of an object moving along the \(s\) -axis as a function of time \(t\). Graph \(f\) together with the velocity function \(v(t)=d s / d t=f^{\prime}(t)\) and the acceleration function \(a(t)=d^{2} s / d t^{2}=f^{\prime \prime}(t) .\) Comment on the object's behavior in relation to the signs and values of \(v\) and \(a .\) Include in your commentary such topics as the following: a. When is the object momentarily at rest? b. When does it move to the left (down) or to the right (up)? c. When does it change direction? d. When does it speed up and slow down? e. When is it moving fastest (highest speed)? Slowest? f. When is it farthest from the axis origin? $$s=t^{2}-3 t+2, \quad 0 \leq t \leq 5$$

Use a CAS to estimate the magnitude of the error in using the linearization in place of the function over a specified interval \(I\). Perform the following steps: a. Plot the function \(f\) over \(I\) b. Find the linearization \(L\) of the function at the point \(a\) c. Plot \(f\) and \(L\) together on a single graph. d. Plot the absolute error \(|f(x)-L(x)|\) over \(I\) and find its maximum value. e. From your graph in part (d), estimate as large a \(\delta>0\) as you can, satisfying \(|x-a|<\delta \quad \Rightarrow \quad|f(x)-L(x)|<\epsilon\) for \(\epsilon=0.5,0.1,\) and \(0.01 .\) Then check graphically to see if your \(\delta\) -estimate holds true. $$f(x)=\sqrt{x}-\sin x, \quad[0,2 \pi], \quad a=2$$
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