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Chapter 2: Problem 17

Find the derivative of the function. $$ g(u)=u \sqrt{1-u^{2}} $$

Short Answer

Expert verified
The derivative of the function \(g(u) = u \sqrt{1 - u^2}\) is \(g'(u) = \sqrt{1 - u^2} - u^2(1 - u^2)^{-\frac{1}{2}}\).

Step by step solution

01

Identify the Functions

The given function can be expressed as the product of two functions, \(f(u) = u \) and \(h(u) = \sqrt{1 - u^2} \).
02

Find the Derivative of \(u \)

The derivative of the function \(f(u) = u \) with respect to \(u \) is \(f'(u) = 1\).
03

Find the Derivative of \(\sqrt{1 - u^2}\)

To find the derivative of the function \(h(u) = \sqrt{1 - u^2} \), we will first rewrite the function as \(h(u) = (1 - u^2)^{\frac{1}{2}} \), and then use the chain rule for derivatives. 1. Let \(y = 1 - u^2\), then \(h(u) = y^{\frac{1}{2}}\). 2. The derivative of \(y^{\frac{1}{2}}\) with respect to \(y \) is \(\frac{1}{2}y^{-\frac{1}{2}}\). 3. The derivative of \(1 - u^2\) with respect to \(u \) is \(-2u\). 4. Multiply the derivatives found in steps 2 and 3: \(-2u \cdot \frac{1}{2}(1 - u^2)^{-\frac{1}{2}}\). So, the derivative of \(h(u) = \sqrt{1 - u^2}\) is \(h'(u) = -u(1 - u^2)^{-\frac{1}{2}}\).
04

Apply the Product Rule

Now we will apply the product rule to find the derivative of \(g(u) = u \sqrt{1 - u^2} \). \(g'(u) = f'(u)h(u) + f(u)h'(u) = 1 \cdot \sqrt{1 - u^2} + u \cdot (-u(1 - u^2)^{-\frac{1}{2}})\)
05

Simplify the Result

Finally, we will simplify the expression for the derivative: \(g'(u) = \sqrt{1 - u^2} - u^2(1 - u^2)^{-\frac{1}{2}}\) So, the derivative of the function \(g(u) = u \sqrt{1 - u^2}\) is \(g'(u) = \sqrt{1 - u^2} - u^2(1 - u^2)^{-\frac{1}{2}}\).

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Key Concepts

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

Product Rule
When it comes to finding derivatives, the product rule is a fundamental technique used for functions that are products of two or more other functions. It states that the derivative of a product of two functions is the derivative of the first function times the second function, plus the first function times the derivative of the second function. In mathematical terms, if we have a function expressed as a product, say, \( y = f(u) \times g(u) \), then its derivative \( y' \) is given by:

\[ y' = f'(u)g(u) + f(u)g'(u) \]
Applying this to the given exercise, where we have \( g(u) = u \times \sqrt{1-u^2} \), we treat \( u \) as \( f(u) \) and \( \sqrt{1-u^2} \) as \( g(u) \), and use the product rule for differentiation as shown in the solution steps.
Chain Rule
The chain rule is another essential concept in calculus, particularly for finding the derivative of a composite function. Essentially, if you have a function within a function, the chain rule allows you to differentiate it correctly. It can be summarized by the statement: the derivative of a composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function.

Consider the exercise where \( h(u) = \sqrt{1 - u^2} \), which is a composition of the square root function and \( 1 - u^2 \). As per the chain rule, we differentiate the outer square root function and multiply it by the derivative of the inner function \( 1 - u^2 \). This is succinctly demonstrated in the steps leading to the derivation of \( h'(u) \).
Implicit Differentiation
Implicit differentiation is used when you have an equation that defines a function implicitly, rather than explicitly. It allows us to find the derivative of a function with respect to a variable, even if the function is not isolated on one side of the equation. This technique is particularly helpful when separating the variables is too complicated or impossible.

The given exercise does not directly involve implicit differentiation, but understanding this concept can be extremely beneficial when dealing with more complex derivatives where functions are not explicitly solved for one variable.
Differentiation Techniques
There are various techniques for differentiating functions beyond the product and chain rules. These include the quotient rule for the division of functions, the power rule for functions to a power, and trigonometric differentiation for functions involving sine, cosine, or other trigonometric functions. Advanced techniques include logarithmic differentiation and differentiating inverse functions.

In the context of our problem, we utilized rewriting the square root in power form to apply the chain rule easily. This transformation is a strategic differentiation technique that simplifies the process and is an excellent example of how flexible and creative you can be when tackling differentiation problems.

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Most popular questions from this chapter

Watching a Rocket Launch At a distance of \(2000 \mathrm{ft}\) from the launch site, a spectator is observing a rocket 120 -ft long being launched vertically. Let \(\theta\) be her viewing angle of the rocket, and let \(y\) denote the altitude (measured in feet) of the rocket. (Neglect the height of the spectator.) a. Show that $$ \tan \theta=\frac{240,000}{y^{2}+120 y+4,000,000} $$ b. What is the viewing angle when the rocket is on the launching pad? When it is at an altitude of 10,000 feet? c. Find the rate of change of the viewing angle when the rocket is at an altitude of 10,000 feet. d. What happens to the viewing angle when the rocket is at a very great altitude?

Orbit of a Satellite An artificial satellite moves around the earth in an elliptic orbit. Its distance \(r\) from the center of the earth is approximated by $$ r=a\left[1-e \cos M-\frac{e^{2}}{2}(\cos 2 M-1)\right] $$ where \(M=(2 \pi / P)\left(t-t_{n}\right) .\) Here, \(t\) is time and \(a, e, P\), and \(t_{n}\) are constants measuring the semimajor axis of the orbit, the eccentricity of the orbit, the period of orbiting, and the time taken by the satellite to pass the perigee, respectively. Find \(d r / d t\), the radial velocity of the satellite.

The magnitude of the gravitational force exerted by the earth on a particle of mass \(m\) at a distance \(r\) from the center of the earth is $$F(r)=\left\\{\begin{array}{ll} \frac{G M m r}{R^{2}} & \text { if } rR\), and interpret your result.

If a right circular cylinder of radius \(a\) is filled with water and rotated about its vertical axis with a constant angular velocity \(\omega\), then the water surface assumes a shape whose cross section in a plane containing the vertical axis is a parabola. If we choose the \(x y\) -system so that the \(y\) -axis is the axis of rotation and the vertex of the parabola passes through the origin of the coordinate system, then the equation of the parabola is $$y=\frac{\omega^{2} x^{2}}{2 g}$$ where \(g\) is the acceleration due to gravity. Find the angle \(\alpha\) that the tangent line to the water level makes with the \(x\) -axis at any point on the water level. What happens to \(\alpha\) as \(\omega\) increases? Interpret your result.

Two families of curves are orthogonal trajectories of each other if every curve of one family is orthogonal to every curve in the other family. In Exercises \(93-96\), (a) show that the given families of curves are orthogonal to each other, and (b) sketch a few members of each family on the same set of axes. $$ x^{2}+y^{2}=c x, \quad x^{2}+y^{2}=k y, \quad c, k \text { constants } $$
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