/*! 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 12 Find \(f^{\prime}(x)\). $$ f... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 12

Find \(f^{\prime}(x)\). $$ f(x)=7 x^{-6}-5 \sqrt{x} $$

Short Answer

Expert verified
The derivative is \(f^{\prime}(x) = -42x^{-7} - \frac{5}{2}x^{-1/2}\).

Step by step solution

01

Identify the Function Components

The function given is \(f(x) = 7x^{-6} - 5\sqrt{x}\). This function has two terms: \(7x^{-6}\) and \(-5\sqrt{x}\). The square root term can be rewritten with a fractional exponent as \(-5x^{1/2}\).
02

Differentiate the First Term

Differentiate the term \(7x^{-6}\) using the power rule. The power rule states that \(\frac{d}{dx}[x^n] = nx^{n-1}\). Thus, \(\frac{d}{dx}[7x^{-6}] = 7 \cdot (-6)x^{-6-1} = -42x^{-7}\).
03

Differentiate the Second Term

Differentiate the term \(-5x^{1/2}\) using the power rule. The derivative is \(-5 \cdot \frac{1}{2} \cdot x^{1/2 - 1} = -\frac{5}{2}x^{-1/2}\).
04

Combine the Derivatives

Add the derivatives obtained from each term to find \(f^{\prime}(x)\). Combine \(-42x^{-7}\) and \(-\frac{5}{2}x^{-1/2}\) to get the derivative of the whole function: \(f^{\prime}(x) = -42x^{-7} - \frac{5}{2}x^{-1/2}\).

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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 technique used in calculus for differentiating functions involving powers of a variable. When you have a function like \(x^n\), the Power Rule allows for easy differentiation. The rule states:
  • The derivative of \(x^n\) is \(n \cdot x^{n-1}\).
To apply this, simply:
  • Take the exponent \(n\) of the variable \(x\).
  • Multiply the entire term by this exponent.
  • Then, reduce the exponent of \(x\) by one.
For example, in the given function \(7x^{-6}\), applying the Power Rule involves multiplying 7 by \(-6\) (the exponent), resulting in \(-42x^{-7}\). This reduces the power by one.
It's a straightforward method that greatly simplifies the process of differentiation for polynomials and other similar expressions.
Fractional Exponents
Understanding fractional exponents is crucial when dealing with roots in calculus. A fractional exponent represents both a power and a root. For instance,
  • \(x^{1/2}\) is equivalent to \(\sqrt{x}\).
  • The expression \(x^{m/n}\) means \(\sqrt[n]{x^m}\).
Rewriting roots using fractional exponents makes it easier to apply the Power Rule and other calculus techniques. Look at \(\sqrt{x}\) in the exercise. Transformed into \(x^{1/2}\), it allows us to handle differentiation using simple exponent manipulation.
Thus, a term like \(5\sqrt{x}\) can be restated as \(5x^{1/2}\). By switching to this format, you can utilize calculus tools more effectively, streamlining the process of finding derivatives and simplifying the work.
Function Differentiation
When differentiating a function with multiple terms, it is important to differentiate each term individually and then combine them. Each term is treated separately, applying the relevant differentiation rules.
  • First, identify and rewrite any roots as fractional exponents.
  • Second, apply the Power Rule and any other necessary rules to each term.
  • Each differentiated term is then summed up to form the derivative of the original function.
In the exercise, \(f(x) = 7x^{-6} - 5\sqrt{x}\) is divided into two terms before differentiation.
By differentiating each part: \(-42x^{-7}\) and \(-\frac{5}{2}x^{-1/2}\), they are combined to form the complete derivative \(f^{\prime}(x) = -42x^{-7} - \frac{5}{2}x^{-1/2}\).
This methodical approach, treating each term step-by-step, ensures accuracy and clarity, especially useful when dealing with more complex functions.

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Most popular questions from this chapter

Determine whether the statement is true or false. Explain your answer. If \(f\) and \(g\) are differentiable at \(x=2\), then $$ \left.\frac{d}{d x}[f(x)-8 g(x)]\right|_{x=2}=f^{\prime}(2)-8 g^{\prime}(2) $$

Suppose that a function \(f\) is differentiable at \(x=0\) with \(f(0)=f^{\prime}(0)=0\), and let \(y=m x, m \neq 0\), denote any line of nonzero slope through the origin. (a) Prove that there exists an open interval containing 0 such that for all nonzero \(x\) in this interval \(|f(x)|<\left|\frac{1}{2} m x\right| .\) [Hint: Let \(\epsilon=\frac{1}{2}|m|\) and apply Definition \(1.4 .1\) to \((5)\) with \(\left.x_{0}=0 .\right]\) (b) Conclude from part (a) and the triangle inequality that there exists an open interval containing 0 such that \(|f(x)|<|f(x)-m x|\) for all \(x\) in this interval. (c) Explain why the result obtained in part (b) may be interpreted to mean that the tangent line to the graph of \(f\) at the origin is the best linear approximation to \(f\) at that point.

If an object suspended from a spring is displaced vertically from its equilibrium position by a small amount and released, and if the air resistance and the mass of the spring are ignored, then the resulting oscillation of the object is called simple harmonic motion. Under appropriate conditions the displacement \(y\) from equilibrium in terms of time \(t\) is given by $$ y=A \cos \omega t $$ where \(A\) is the initial displacement at time \(t=0\), and \(\omega\) is a constant that depends on the mass of the object and the stiffness of the spring (see the accompanying figure). The constant \(|A|\) is called the amplitude of the motion and \(\omega\) the angular frequency. (a) Show that $$ \frac{d^{2} y}{d t^{2}}=-\omega^{2} y $$ (b) The period \(T\) is the time required to make one complete oscillation. Show that \(T=2 \pi / \omega\). (c) The frequency \(f\) of the vibration is the number of oscillations per unit time. Find \(f\) in terms of the period \(T\). (d) Find the amplitude, period, and frequency of an object that is executing simple harmonic motion given by \(y=0.6 \cos 15 t\), where \(t\) is in seconds and \(y\) is in centimeters.

Determine whether the statement is true or false. Explain your answer. $$ \begin{aligned} &\text { If } f^{\prime}(2)=5 \text { , then } \\ &\left.\frac{d}{d x}\left[4 f(x)+x^{3}\right]\right|_{x=2}=\left.\frac{d}{d x}[4 f(x)+8]\right|_{x=2}=4 f^{\prime}(2)=20 \end{aligned} $$

Show that if \(x \neq 0\), then \(y=1 / x\) satisfies the equation \(x^{3} y^{\prime \prime}+x^{2} y^{\prime}-x y=0\)
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