/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 30 \(f(x)=x^{-1}, \quad x_{0}=1\)... [FREE SOLUTION] | 91Ó°ÊÓ

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\(f(x)=x^{-1}, \quad x_{0}=1\)

Short Answer

Expert verified
The derivative of \(f(x) = x^{-1}\) at \(x_{0} = 1\) is -1.

Step by step solution

01

Rewrite the function

We first rewrite the function \(f(x)\) in its power form as \(f(x) = x^{-1}\). This is already done for us.
02

Apply the power rule

The power rule for differentiation states that if \(f(x) = x^n\), then \(f'(x) = n*x^{(n-1)}\). Applying this rule to our function, we find the derivative \(f'(x) = -1 * x^{-2} = -x^{-2}\).
03

Evaluate at the given point

Now that we have the derivative, we plug in the given point \(x_{0}=1\) into \(f'(x)\) to get \(f'(1) = -1 * 1^{-2} = -1\).

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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 differential calculus. It simplifies finding the derivative of functions that are expressed as powers of x. For any function of the form \(f(x) = x^n\), the Power Rule states that its derivative is \(f'(x) = n \cdot x^{n-1}\). This means you multiply the entire expression by the exponent \(n\) and then subtract one from the exponent.
For example:
  • If you have \(x^3\), applying the Power Rule gives the derivative \(3x^{2}\).
  • For \(x^{-1}\), the derivative becomes \(-1x^{-2}\), as illustrated in our exercise.
The Power Rule makes it easy to compute derivatives for polynomial functions, which are quite common and usually follow simple forms. Remember, the key is recognizing the exponent and knowing how to adjust it through differentiation.
Derivative Evaluation
Once we have a derivative, we can evaluate it at specific points. This is particularly useful for understanding the behavior of a function at a particular value of x. Evaluating a derivative requires substituting the given x-value into the derivative function.
Consider our example: after applying the Power Rule to \(f(x) = x^{-1}\), we find that \(f'(x) = -x^{-2}\). By evaluating this at \(x_0 = 1\), we plug 1 into the derivative and calculate \(f'(1) = -1 \cdot 1^{-2} = -1\).
This process tells us the instantaneous rate of change of the function at \(x = 1\). Evaluating derivatives is a step that connects the general rule we've developed (in this case, the derivative expression) to practical insights about the function's specific behavior.
Function Derivatives
Function derivatives are at the heart of understanding how functions change. Calculating these involves using rules such as the Power Rule and others, like the Product or Chain Rule, depending on the function's complexity.
For a simple function like \(f(x) = x^{-1}\), we apply the Power Rule to derive its changes as \(f'(x) = -x^{-2}\).
  • The negative sign indicates the function is decreasing.
  • The term \(x^{-2}\) shows the rate at which the function decreases reduces as x increases.
Understanding function derivatives is crucial. They help predict and explain the changes observed in functions, laying groundwork for further calculus concepts such as integrals and more comprehensive modeling of dynamic situations.

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Most popular questions from this chapter

$$x^{2} y^{\prime \prime}+x y^{\prime}+x^{2} y=0$$

Buckling Columns. In the study of the buckling of a column whose cross section varies, one encounters the equation $$\quad x^{n} y^{\prime \prime}(x)+\alpha^{2} y(x)=0, \quad x>0$$ where x is related to the height above the ground and y is the deflection away from the vertical. The positive constant a depends on the rigidity of the column, its moment of inertia at the top, and the load. The positive integer n depends on the type of column. For example, when the column is a truncated cone [see Figure 8.13(a) on page 474], we have $$n=4$$ (a) Use the substitution \(x=t^{-1}\) to reduce \((45)\) with \(n=4\) to the form \(\frac{d^{2} y}{d t^{2}}+\frac{2}{t} \frac{d y}{d t}+\alpha^{2} y=0, \quad t>0\) (b) Find at least the first six nonzero terms in the series expansion about \(t=0\) for a general solution to the equation obtained in part (a). (c) Use the result of part (b) to give an expansion about \(x=\infty\) for a general solution to \((45) .\)

(a) Construct the Taylor polynomial \(p_{3}(x)\) of degree 3 for the function \(f(x)=1 /(2-x)\) around \(x=0\) $$$$(b) Using the error formula (6), show that $$ \left|f\left(\frac{1}{2}\right)-p_{3}\left(\frac{1}{2}\right)\right|=\left|\frac{2}{3}-p_{3}\left(\frac{1}{2}\right)\right| \leq \frac{2}{3^{5}}$$ $$$$(c) Compare the estimate in part (b) with the actual error $$ \left|\frac{2}{3}-p_{3}\left(\frac{1}{2}\right)\right| $$ $$$$(d) Sketch the graphs of \(1 /(2-x)\) and \(p_{3}(x)\) (on the same axes) for \(-2\)<\(x\)<\(2\)

\(\left(1-x^{2}\right) y^{\prime \prime}-y^{\prime}+y=\tan x\)

\(9 x^{2} y^{\prime \prime}+9 x y^{\prime}+\left(9 x^{2}-16\right) y=0\)

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