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Chapter 3: Problem 49

In each of Exercises \(45-50\) use the Chain Rule repeatedly to determine the derivative with respect to \(x\) of the given expression. $$ 2 \tan (\sqrt{3 x+2}) $$

Short Answer

Expert verified
The derivative of \(2 \tan (\sqrt{3x+2})\) with respect to \(x\) is \(3 \sec^2(\sqrt{3x+2}) \cdot \frac{1}{\sqrt{3x+2}}\).

Step by step solution

01

Identify the Function Layers

The function we need to differentiate is \(2 \tan (\sqrt{3x+2})\). This is a composition of three functions: the innermost function \(u = 3x+2\), the square root function \(v = \sqrt{u}\), and the tangent function \(w = \tan(v)\). The entire expression is then \(2w\).
02

Differentiate the Outermost Layer

Start by differentiating the outermost layer, which is \(2w\), in terms of \(w\), giving \(\frac{d}{dw}[2w] = 2\cdot \frac{d}{dw}[w] = 2\).
03

Differentiate the Tangent Layer

Next, differentiate \(w = \tan(v)\) with respect to \(v\). The derivative of \(\tan(v)\) is \(\sec^2(v)\).
04

Differentiate the Square Root Layer

Differentiate the function \(v = \sqrt{u}\) where \(u = 3x+2\). The derivative is \(\frac{d}{du}[\sqrt{u}] = \frac{1}{2\sqrt{u}}\).
05

Differentiate the Innermost Layer

Differentiate \(u = 3x + 2\) with respect to \(x\), which gives \(\frac{d}{dx}[3x+2] = 3\).
06

Apply the Chain Rule

Combine all derivatives using the chain rule: \(\frac{dy}{dx} = \frac{dy}{dw} \cdot \frac{dw}{dv} \cdot \frac{dv}{du} \cdot \frac{du}{dx}\). Substituting the derivatives, we get: \(2 \cdot \sec^2(v) \cdot \frac{1}{2\sqrt{u}} \cdot 3\).
07

Substitute Back the Variables

Substitute back \(u = 3x+2\) and \(v = \sqrt{3x+2}\) into the chain rule expression: \(\frac{dy}{dx} = 3 \cdot \sec^2(\sqrt{3x+2}) \cdot \frac{1}{\sqrt{3x+2}}\).

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Key Concepts

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

Derivative
In calculus, the concept of a derivative is crucial. It represents the rate at which a function is changing at any given point. Think of it as a way to understand how one quantity changes in response to a change in another. When you're driving a car, the speedometer shows your speed at that exact moment 鈥� that's akin to a derivative, showing your rate of change.
  • The derivative of a function is often denoted as \( f'(x) \) or \( \frac{dy}{dx} \), indicating how the function \( y \) changes as the input \( x \) changes.
  • To find the derivative, we apply various rules of differentiation, such as the Chain Rule.
In our original exercise, we aim to find the derivative of the expression \( 2 \tan(\sqrt{3x+2}) \). Understanding how to calculate derivatives is fundamental for tackling many problems in physics, engineering, and beyond.
Composition of Functions
Functions can be combined in various ways, and one important method is through composition. When you have two functions, say \( f(x) \) and \( g(x) \), you can form a new function by plugging one into the other. This is called a composite function, denoted as \( f(g(x)) \).
  • The outer function \( f \) acts on whatever \( g(x) \) outputs. This repeated application process is fundamental to solving certain problems.
  • With composite functions, the Chain Rule helps us compute derivatives.
In this exercise, we see composition through several layers: \( 3x+2 \) forms the base, \( \sqrt{3x+2} \) is the next layer, and \( \tan(\sqrt{3x+2}) \) is another layer inside the function \( 2\tan(v) \). Each layer is seamlessly incorporated into the next.
Tangent Function
The tangent function, expressed as \( \tan(x) \), is a trigonometric function that represents the ratio of the opposite to the adjacent side of a right triangle. It's an essential part of trigonometry and calculus.
  • The tangent function is periodic, with a period of \( \pi \), meaning it repeats every \( \pi \) units.
  • Its derivative, \( \sec^2(x) \), plays a critical role in finding the rate of change of tangent functions in calculus problems.
In our problem, we work with \( \tan(\sqrt{3x+2}) \). Deriving the tangent function results in \( \sec^2(v) \) where \( v = \sqrt{3x+2} \). This showcases how trigonometric functions can be differentiated using the Chain Rule.
Square Root Function
The square root function, denoted as \( \sqrt{x} \), is commonly encountered in mathematics. It signifies a number which, when multiplied by itself, results in \( x \). Understanding its derivative is crucial for analyzing many mathematical models.
  • The derivative of \( \sqrt{x} \) is \( \frac{1}{2\sqrt{x}} \), which shows how the function \( \sqrt{x} \) changes as \( x \) changes.
  • This derivative becomes useful when the square root function is part of a larger expression, especially in composite functions.
In our exercise, the square root function is used in \( \sqrt{3x+2} \). By differentiating it, we find that \( \frac{1}{2\sqrt{u}} \) where \( u = 3x+2 \). This step is integral to piecing together our complete derivative using the Chain Rule.

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Most popular questions from this chapter

Use a central difference quotient to approximate \(f^{\prime}(c)\) for the given \(f\) and \(c .\) Plot the function and the tangent line at \((c, f(c))\). $$ f(x)=\sinh \left(\frac{x}{1+\sqrt{x}}\right), \quad c=4.5 $$

Calculate the value of the given inverse trigonometric function at the given point. $$ \arccos (-\sqrt{3} / 2) $$

Calculate the linearization \(L(x)=f(c)+\) \(f^{\prime}(c), \cdot(x-c)\) for the given function \(f\) at the given value \(c\) $$ f(x)=\cos (x), c=\pi / 3 $$

An initial value problem is given, along with its exact solution. (Read the instructions for Exercises \(47-50\) for terminology.) Verify that the given solution is correct by substituting it into the given differential equation and the initial value condition. Calculate the Euler's Method approximation \(y_{1}=y_{0}+F\left(x_{0}, y_{0}\right) \Delta x\) of \(y\left(x_{1}\right)\) where \(\Delta x=x_{1}-x_{0} .\) Let \(m_{1}=\left(F\left(x_{0}, y_{0}\right)+F\left(x_{1}, y_{1}\right)\right) / 2\) and \(z_{1}=y_{0}+\) \(m_{1} \Delta x .\) This is the Improved Euler Method approximation of \(y\left(x_{1}\right) .\) Calculate \(z_{1} .\) By evaluating \(y\left(x_{1}\right),\) determine which of the two approximations, \(y_{1}\) or \(z_{1},\) is more accurate. $$ \begin{array}{l} d y / d x=x^{2}-2 y, y(0)=3, x_{1}=1 / 4 ; \text { Exact solution: } y(x)= \\ x^{2} / 2-x / 2+1 / 4+11 / 4 \exp (-2 x) \end{array} $$

Show that \(\arctan (x)=\operatorname{arccot}(1 / x)\) for all \(x>0\) and \(\arctan (x)=\operatorname{arccot}(1 / x)-\pi\) for all \(x<0\).
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