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Chapter 3: Problem 18

In Exercises \(1-18,\) find \(d y / d x\) $$ g(x)=(2-x) \tan ^{2} x $$

Short Answer

Expert verified
The derivative of \( g(x) = (2-x) \tan^2 x \) is \( -\tan^2 x + 2\tan x \sec^2 x - x\tan x \sec^2 x \).

Step by step solution

01

Understand the function

The function given is \( g(x) = (2-x) \tan^2 x \). We need to find the derivative \( \frac{d}{dx} g(x) \). This function is the product of two functions: \( u(x) = 2-x \) and \( v(x) = \tan^2 x \).
02

Apply the Product Rule

We use the product rule for differentiation, which states \( \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \). Identify \( u(x) = 2-x \) and \( v(x) = \tan^2 x \). First, find \( u'(x) \) and \( v'(x) \).
03

Differentiate \( u(x) \)

The function \( u(x) = 2-x \) is a simple linear function. Its derivative is \( u'(x) = -1 \).
04

Differentiate \( v(x) \) using the Chain Rule

To find \( v'(x) \), where \( v(x) = \tan^2 x \), use the chain rule. Let \( w(x) = \tan x \) then \( v(x) = w(x)^2 \). Differentiate to get \( v'(x) = 2w(x)w'(x) = 2\tan x \sec^2 x \) since \( w'(x) = \sec^2 x \).
05

Substitute and Simplify

Substitute \( u(x) = 2-x \), \( u'(x) = -1 \), \( v(x) = \tan^2 x \), and \( v'(x) = 2\tan x \sec^2 x \) into the product rule formula: \( \frac{d}{dx}[ (2-x)\tan^2 x ] = (-1)\tan^2 x + (2-x)(2\tan x \sec^2 x) \). Simplify the expression to get the final derivative.

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Key Concepts

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

Product Rule
When dealing with derivatives, especially of functions that are products of two or more distinct parts, we use the product rule. The product rule is a formula that helps us differentiate an expression where two functions are multiplied together. For any two functions, say \( u(x) \) and \( v(x) \), their product \( u(x) \cdot v(x) \) has a derivative given by:
  • \( \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \)
This says that the derivative of the product is the derivative of the first function times the second function plus the first function times the derivative of the second function.
In our problem, the functions were \( u(x) = 2-x \) and \( v(x) = \tan^2 x \).
Knowing this rule allows us to break down the entire process into simpler parts.
Chain Rule
The chain rule is another fundamental tool in calculus, particularly useful when dealing with complex functions composed of other functions. It lets us differentiate a function that is nested within another. For a composite function \( f(g(x)) \), the derivative is:
  • \( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \)
In this context, \( v(x) = \tan^2 x \) is one such composite function because it can be rewritten as \( (\tan x)^2 \).
Here we apply the chain rule by first considering \( w(x) = \tan x \) so that \( v(x) = w(x)^2 \). We differentiate to find \( v'(x) \): by finding the derivative \( w'(x) = \sec^2 x \) and using the rule \( 2w(x)w'(x) \), resulting in \( 2\tan x \sec^2 x \).
Trigonometric Derivatives
Understanding trigonometric derivatives is crucial in calculus when working with functions involving trigonometric expressions like sine, cosine, or tangent. Each of these functions has a well-known derivative:
  • \( \frac{d}{dx} [\sin x] = \cos x \)
  • \( \frac{d}{dx} [\cos x] = -\sin x \)
  • \( \frac{d}{dx} [\tan x] = \sec^2 x \)
In the original exercise, these formulas came into play when determining \( v'(x) \) since \( \tan x \) appears in the function \( v(x) = \tan^2 x \).
Recognizing that the derivative of \( \tan x \) is \( \sec^2 x \) is key to correctly applying both the chain rule and product rule in the process.

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Most popular questions from this chapter

If \(r+s^{2}+v^{3}=12, d r / d t=4,\) and \(d s / d t=-3,\) find \(d v / d t\) when \(r=3\) and \(s=1.\)

Find \(d y / d x\) if \(y=x^{3 / 2}\) by using the Chain Rule with \(y\) as a composite of $$\begin{array}{l}{\text { a. } y=u^{3} \text { and } u=\sqrt{x}} \\ {\text { b. } y=\sqrt{u} \text { and } u=x^{3}}\end{array}$$

The effect of flight maneuvers on the heart The amount of work done by the heart's main pumping chamber, the left ventricle, is given by the equation $$ W=P V+\frac{V \delta v^{2}}{2 g} $$ where \(W\) is the work per unit time, \(P\) is the average blood pressure, \(V\) is the volume of blood pumped out during the unit of time, \(\delta(\) "delta") is the weight density of the blood, \(v\) is the average velocity of the exiting blood, and \(g\) is the acceleration of gravity. $$ \begin{array}{l}{\text { When } P, V, \delta, \text { and } v \text { remain constant, } W \text { becomes a function }} \\ {\text { of } g, \text { and the equation takes the simplified form }}\end{array} $$ $$ W=a+\frac{b}{g}(a, b \text { constant }) $$ As a member of NASA's medical team, you want to know how sensitive \(W\) is to apparent changes in \(g\) caused by flight maneuvers, and this depends on the initial value of \(g\) . As part of your investigation, you decide to compare the effect on \(W\) of a given change \(d g\) on the moon, where \(g=5.2 \mathrm{ft} / \mathrm{sec}^{2},\) with the effect the same change \(d g\) would have on Earth, where \(g=32 \mathrm{ft} / \mathrm{sec}^{2} .\) Use the simplified equation above to find the ratio of \(d W_{\mathrm{moon}}\) to \(d W_{\mathrm{Earth}}\)

In Exercises \(57-60,\) use a CAS to estimate the magnitude of the error in using the linearization in place of the function over a specified interval I. Perform the following steps: $$ \begin{array}{l}{\text { a. Plot the function } f \text { over } I} \\ {\text { b. Find the linearization } L \text { of the function at the point } a \text { . }} \\ {\text { c. Plot } f \text { and } L \text { together on a single graph. }} \\ {\text { d. Plot the absolute error }|f(x)-L(x)| \text { over } I \text { and find its max- }} \\ {\text { imum value. }}\end{array} $$ $$ \begin{array}{l}{\text { e. From your graph in part (d), estimate as large a } \delta>0 \text { as you }} \\ {\text { can, satisfing }}\end{array} $$ $$ \begin{array}{c}{|x-a|<\delta \quad \Rightarrow \quad|f(x)-L(x)|<\epsilon} \\\ {\text { for } \epsilon=0.5,0.1, \text { and } 0.01 . \text { Then check graphically to see if }} \\ {\text { your } \delta \text { -estimate holds true. }}\end{array} $$ $$ f(x)=x^{2 / 3}(x-2), \quad[-2,3], \quad a=2 $$

The linearization is the best linear approximation Suppose that \(y=f(x)\) is differentiable at \(x=a\) and that \(g(x)=\) \(m(x-a)+c\) is a linear function in which \(m\) and \(c\) are constants. If the error \(E(x)=f(x)-g(x)\) were small enough near \(x=a\) we might think of using \(g\) as a linear approximation of \(f\) instead of the linearization \(L(x)=f(a)+f^{\prime}(a)(x-a) .\) Show that if we impose on \(g\) the conditions $$ \begin{array}{l}{\text { 1. } E(a)=0} \\ {\text { 2. } \lim _{x \rightarrow a} \frac{E(x)}{x-a}=0}\end{array} $$ then \(g(x)=f(a)+f^{\prime}(a)(x-a) .\) Thus, the linearization \(L(x)\) gives the only linear approximation whose error is both zero at \(x=a\) and negligible in comparison with \(x-a\) .
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