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Chapter 3: Problem 2

Differentiate. \(f(x)=\sqrt{x} \sin x\)

Short Answer

Expert verified
The derivative is \( f'(x) = \frac{\sin x}{2\sqrt{x}} + \cos x \sqrt{x} \).

Step by step solution

01

Identify the Rule to Use

The function to differentiate is a product of two functions: \( u(x) = \sqrt{x} \) and \( v(x) = \sin x \). Thus, we will use the product rule for differentiation: \( (uv)' = u'v + uv' \).
02

Differentiate \(u(x)\)

First, find the derivative of \( u(x) = \sqrt{x} = x^{1/2} \). Using the power rule, \( u'(x) = \frac{1}{2}x^{-1/2} \).
03

Differentiate \(v(x)\)

The function \( v(x) = \sin x \). The derivative of \( \sin x \) is \( \cos x \), so \( v'(x) = \cos x \).
04

Apply the Product Rule

Using the product rule, substitute the derivatives into the formula: \( (uv)' = u'v + uv' \).\[ f'(x) = \frac{1}{2}x^{-1/2} \cdot \sin x + \sqrt{x} \cdot \cos x \]
05

Simplify the Expression

Combine and simplify the expression for the derivative: \[ f'(x) = \frac{1}{2} \cdot \frac{\sin x}{\sqrt{x}} + \cos x \sqrt{x} \]. This can be rewritten as \[ f'(x) = \frac{\sin x}{2\sqrt{x}} + \cos x \sqrt{x} \].

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Key Concepts

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

Power Rule Differentiation
The power rule is a fundamental tool in calculus for differentiating polynomial-like expressions. If you have a function of the form \( f(x) = x^n \), where \( n \) is a constant, the power rule states that the derivative is \( f'(x) = nx^{n-1} \). This is incredibly useful because it simplifies finding derivatives of expressions like \( x^2 \) or \( x^{1/2} \), the latter being the square root of x.

To illustrate, consider the function \( u(x) = \sqrt{x} = x^{1/2} \). Applying the power rule, we replace \( n \) with \( 1/2 \), giving us:
  • First, bring down the exponent as a coefficient: \( \frac{1}{2} \).
  • Then, subtract 1 from the exponent: \( \frac{1}{2} - 1 = -\frac{1}{2} \).
  • Thus, the derivative is \( u'(x) = \frac{1}{2}x^{-1/2} \), which can also be written as \( \frac{1}{2} \frac{1}{\sqrt{x}} \).
By understanding these steps, the power rule becomes simple to apply.
Derivative of Sine Function
The sine function is one of the primary trigonometric functions, and its derivative is straightforward due to its cyclic nature. For the function \( v(x) = \sin x \), the derivative is \( v'(x) = \cos x \). This derivative comes from understanding the unit circle and how the sine and cosine functions represent projections of a rotating vector.

Here's how you can think about it:
  • As you move around the unit circle, the sine value (y-coordinate) changes smoothly.
  • The cosine (x-coordinate) represents the slope or rate of change of the sine function.
  • Hence, at any point \( x \), the rate at which \( \sin x \) is changing is given by \( \cos x \).
This fundamental property allows us to quickly find the derivative of functions involving sine, which is immensely useful in both pure mathematics and applied scenarios.
Differentiating Square Roots
Differentiating square roots may initially seem complex, but it becomes much simpler with the power rule. When confronted with \( \sqrt{x} \), rewrite it as \( x^{1/2} \) to make applying calculus rules straightforward.

Here's a step-by-step approach:
  • Rewrite \( \sqrt{x} \) as \( x^{1/2} \), making it compatible with the power rule.
  • Apply the power rule: Bring down \( 1/2 \) and reduce the exponent by 1 to get \( -1/2 \).
  • The derivative \( u'(x) = \frac{1}{2}x^{-1/2} \) is equivalent to \( \frac{1}{2\sqrt{x}} \).
By converting square roots to exponents, differentiation becomes more consistent and straightforward, allowing for easier integration into more complex problems, such as using the product rule.

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Most popular questions from this chapter

Find \(y^{\prime}\) if \(x^{y}=y^{x}\)

If \(R\) denotes the reaction of the body to some stimulus of strength \(x,\) the sensitivity \(S\) is defined to be the rate of change of the reaction with respect to \(x\) . A particular example is that when the brightness \(x\) of a light source is increased, the eye reacts by decreasing the area \(R\) of the pupil. The experimental formula $$R=\frac{40+24 x^{0.4}}{1+4 x^{0.4}}$$ has been used to model the dependence of \(\mathrm{R}\) on \(\mathrm{x}\) when \(\mathrm{R}\) is measured in square millimeters and \(\mathrm{x}\) is measured in appropriate units of brightness. (a) Find the sensitivity. (b) Illustrate part (a) by graphing both \(\mathrm{R}\) at low levels of of \(\mathrm{x} .\) Comment on the values of \(\mathrm{R}\) and \(\mathrm{S}\) at low levels of brightness. Is this what you would expect?

A television camera is positioned 4000 \(\mathrm{ft}\) from the base of a rocket launching pad. The angle of elevation of the camera has to change at the correct rate in order to keep the rocket in sight. Also, the mechanism for focusing the camera has to take into account the increasing distance from the camera to the rising rocket. Let's assume the rocket rises vertically and its speed is 600 \(\mathrm{ft} / \mathrm{s}\) when it has risen 3000 \(\mathrm{ft}\) . (a) How fast is the distance from the television camera to the rocket changing at that moment? (b) If the television camera is always kept aimed at the rocket, how fast is the camera's angle of elevation changing at that same moment?

Brain weight \(\mathrm{B}\) as a function of body weight \(\mathrm{W}\) in fish has been modeled by the power function \(\mathrm{B}=0.007 \mathrm{W}^{2 / 3}\) , where \(\mathrm{B}\) and \(\mathrm{W}\) are measured in grams. A model for body weight as a function of body length \(\mathrm{L}\) (measured in centimeters) is \(\mathrm{W}=0.12 \mathrm{L}^{2.53}\) . If, over 10 million years, the average length of a certain species of fish evolved from 15 \(\mathrm{cm}\) to 20 \(\mathrm{cm}\) at a constant rate, how fast was this species' brain growing when the average length was 18 \(\mathrm{cm} ?\)

If two resistors with resistances \(\mathrm{R}_{1}\) and \(\mathrm{R}_{2}\) are connected in parallel, as in the figure, then the total resistance \(\mathrm{R}\) , measured in in ohms \((\Omega),\) is given by $$\frac{1}{\mathrm{R}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}$$ If \(R_{1}\) and \(R_{2}\) are increasing at rates of 0.3\(\Omega / s\) and 0.2\(\Omega / s\) respectively, how fast is \(R\) changing when \(R_{1}=80 \Omega\) and \(R_{2}=100 \Omega ?\)
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