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Chapter 2: Problem 41

Find \(f^{\prime}(x)\) by using the definition of the derivative. $$f(x)=\sqrt{x}$$

Short Answer

Expert verified
The derivative \(f'(x) = \frac{1}{2\sqrt{x}}\).

Step by step solution

01

Understand the Problem

We need to find the derivative of the function \(f(x) = \sqrt{x}\) using the definition of the derivative. The definition of the derivative \(f'(x)\) at a point \(x\) is given by:\[f'(x) = \lim_{{h \to 0}} \frac{{f(x+h) - f(x)}}{h}.\]
02

Substitute into the Derivative Formula

Substitute \(f(x) = \sqrt{x}\) into the limit definition of the derivative:\[f'(x) = \lim_{{h \to 0}} \frac{{\sqrt{x+h} - \sqrt{x}}}{h}.\]
03

Rationalize the Numerator

To simplify the expression, multiply the numerator and the denominator by the conjugate of the numerator:\[f'(x) = \lim_{{h \to 0}} \frac{{\sqrt{x+h} - \sqrt{x}}} {h} \times \frac{{\sqrt{x+h} + \sqrt{x}}}{\sqrt{x+h} + \sqrt{x}} = \lim_{{h \to 0}} \frac{{(\sqrt{x+h})^2 - (\sqrt{x})^2}}{h(\sqrt{x+h} + \sqrt{x})}.\]
04

Simplify the Expression

In the expression \((\sqrt{x+h})^2 - (\sqrt{x})^2\), use the difference of squares formula:\[(x+h) - x = h.\]Then, the fraction in the limit becomes:\[f'(x) = \lim_{{h \to 0}} \frac{h}{h(\sqrt{x+h} + \sqrt{x})}.\]
05

Cancel \(h\) and Evaluate the Limit

Cancel \(h\) in the numerator and denominator:\[f'(x) = \lim_{{h \to 0}} \frac{1}{\sqrt{x+h} + \sqrt{x}}.\]Now evaluate the limit as \(h\) approaches 0:\[f'(x) = \frac{1}{2\sqrt{x}}.\]
06

Conclusion

We used the limit definition of the derivative to find that the derivative of \(f(x) = \sqrt{x}\) is:\[f'(x) = \frac{1}{2\sqrt{x}}.\]

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Key Concepts

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

Derivative
The derivative is a fundamental concept in calculus. It represents the rate at which a function changes at any given point. To visualize this, imagine drawing a tangent line to a curve at a point; the slope of this line is the derivative at that point.
In mathematical terms, the derivative of a function \( f(x) \) is denoted as \( f^{\prime}(x) \). This derivative tells us how \( f(x) \) changes with small changes in \( x \).
  • For example, if \( f(x) = \sqrt{x} \), the derivative \( f'(x) \) will tell us how the square root function changes as \( x \) changes.
  • Understanding derivatives is crucial for optimizing functions, finding maximum or minimum values, and analyzing rates of change in real-world situations.
Derivatives are used across science and engineering to describe motion, growth, and many other dynamic phenomena.
Limit Definition
The definition of a derivative relies on the concept of limits, an essential idea in calculus. A limit helps us understand the behavior of a function as the input approaches a specific value. In terms of derivatives, we use limits to find how a function's output changes as the input gets infinitesimally close to a particular point.
To define a derivative using limits, we use the formula:
\[ f^{\prime}(x) = \lim_{{h \to 0}} \frac{{f(x+h) - f(x)}}{h} \]
  • This formula expresses the idea of a function's change over an extremely small interval \( h \) as \( h \) approaches zero.
  • The expression \( \frac{f(x+h) - f(x)}{h} \) represents the average rate of change of the function across an interval \( h \).
Once \( h \) gets close to zero, this average rate of change essentially becomes the instantaneous rate of change, giving us the derivative.
Rationalizing the Numerator
Rationalizing the numerator is a technique used to simplify expressions that include square roots, particularly when applying the limit definition of a derivative.
This process involves multiplying the numerator and the denominator of a fraction by the conjugate of the numerator.
  • When we have an expression like \( \frac{\sqrt{x+h} - \sqrt{x}}{h} \), we multiply by \( \frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}} \) to eliminate the square roots in the numerator.
  • This helps in simplifying the expression, especially when it comes to finding the limit as \( h \) approaches zero.
This simplification is crucial in progressing towards a clear evaluation of the limit, finally obtaining the derivative in a more manageable form.
Difference of Squares
The difference of squares is an algebraic identity that states \((a^2 - b^2) = (a-b)(a+b)\). This formula is useful in calculus, particularly when simplifying expressions in the context of derivatives.
In our original problem, after rationalizing the numerator, we encounter a perfect opportunity to use the difference of squares:
  • We express \((\sqrt{x+h})^2 - (\sqrt{x})^2\) as \((x+h) - x\), which simplifies directly to \( h \).
  • This simplification allows us to cancel \( h \) from the numerator and denominator, making evaluating the limit straightforward.
Using the difference of squares helps reduce complexities in an expression, leading to clearer solutions and making derivations more efficient.

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Most popular questions from this chapter

a. Show that the definition of the derivative applied to the function \(f(x)=\sqrt{x}\) at \(x=0\) gives \(f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{\sqrt{h}}{h}\). b. Use a calculator to evaluate the difference quotient \(\frac{\sqrt{h}}{h}\) for the following values of \(h\) : \(0.1,0.001\), and \(0.00001\). [Hint: Enter the calculation into your calculator with \(h\) replaced by \(0.1\), and then change the value of \(h\) by inserting zeros.] c. From your answers to part (b), does the limit exist? Does the derivative of \(f(x)=\sqrt{x}\) at \(x=0\) exist? d. Graph \(f(x)=\sqrt{x}\) on the window \([0,1]\) by \([0,1]\). Do you see why the slope at \(x=0\) does not exist?

A rocket can rise to a height of \(h(t)=t^{3}+0.5 t^{2}\) feet in \(t\) seconds. Find its velocity and acceleration 10 seconds after it is launched.

Using only straight lines, sketch a function that (a) is continuous everywhere and (b) is differentiable everywhere except at \(x=1\) and \(x=3\).

Graph the cost function \(y_{1}=\sqrt{4 x^{2}+900}\) on the window \([0,30]\) by \([-10,70]\). Then use NDERIV to define \(y_{2}\) as the derivative of \(y_{1}\). Verify the answer to Exercise 53 by evaluating the marginal cost function \(y_{2}\) at \(x=20\).

Use the Generalized Power Rule to find the derivative of each function. $$f(x)=\sqrt{x^{6}+3 x-1}$$
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