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Chapter 6: Problem 2

Find the third order derivative of the function \(f(x)=x e^{2 x}\), \(x \in \mathbb{R}\). In the last section we found the derived functions for sin, cos and exp. We now ask you to find the derived functions for the remaining trigonometric functions and the three most common hyperbolic functions.

Short Answer

Expert verified
The third derivative of \(f(x) = xe^{2x}\) is \( f'''(x) = 16e^{2x} + 8xe^{2x} \).

Step by step solution

01

Identify Necessary Differentiation Rule

To find the derivative of the function \( f(x) = x e^{2x} \), we must apply the product rule of differentiation, which states that the derivative of two functions \( u(x)\) and \( v(x)\) is given by \( (uv)' = u'v + uv' \). Here, we will repeatedly apply this rule to find higher-order derivatives.
02

Apply First Derivative Using Product Rule

Define \( u(x) = x \) and \( v(x) = e^{2x} \). Compute \( u'(x) = 1 \) and \( v'(x) = 2e^{2x} \). Using the product rule, the first derivative is \( f'(x) = 1 \cdot e^{2x} + x \cdot 2e^{2x} = e^{2x} + 2xe^{2x} \).
03

Compute Second Derivative

Differentiate the result from Step 2: \( f'(x) = e^{2x} + 2xe^{2x} \). Apply the product rule again for \( 2xe^{2x} \). Compute \((2xe^{2x})' = 2(e^{2x} + 2xe^{2x}) = 2e^{2x} + 4xe^{2x}\). Thus, \( f''(x) = (e^{2x})' + (2xe^{2x})' = 2e^{2x} + 2e^{2x} + 4xe^{2x} = 4e^{2x} + 4xe^{2x} \).
04

Compute Third Derivative

Differentiate \( f''(x) = 4e^{2x} + 4xe^{2x} \). Again, apply the product rule to \( 4xe^{2x} \). We have \( 4(e^{2x} + 2xe^{2x}) = 4e^{2x} + 8xe^{2x} \). Combine this with the derivative of the constant term: \( f'''(x) = 8e^{2x} + 8e^{2x} + 8xe^{2x} = 16e^{2x} + 8xe^{2x} \).

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Key Concepts

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

Product Rule
When finding the derivative of a product of two functions, you need to use the product rule.
This is a fundamental rule in differentiation that simplifies the process. It states: if you have two functions, written as \( u(x) \) and \( v(x) \), their derivative, \( (uv)' \), is given by \( u'v + uv' \).
Here's a quick breakdown:
  • Differentiate the first function \( u(x) \): The derivative is \( u'(x) \).
  • Leave the second function \( v(x) \) unchanged: This is \( v(x) \).
  • Multiply them: This forms the first part, \( u'(x)v(x) \).
  • Then, differentiate the second function \( v(x) \): The derivative is \( v'(x) \).
  • Leave the first function \( u(x) \) unchanged: This is \( u(x) \).
  • Multiply these two: This is the second part, \( u(x)v'(x) \).
Combine both parts to get the full product rule: \( (uv)' = u'v + uv' \). This rule is crucial when dealing with more complex functions, especially when seeking higher-order derivatives.
Higher-order Derivatives
A higher-order derivative is just a derivative taken multiple times over a function.
The first derivative \( f'(x) \) tells you the rate of change of the function, like velocity in physics.
You find the second derivative \( f''(x) \) by differentiating the first derivative. This second derivative can indicate the function's concavity or its graph's 'curvature'. Think of it as the acceleration of a moving object.
  • First derivative: Rate of change or speed.
  • Second derivative: Acceleration or how the rate is changing.
  • Third derivative and beyond: For most practical purposes, these indicate more subtle changes like the jerk or the rate of change of acceleration.
For the given function \( f(x) = xe^{2x} \), you apply the product rule repeatedly. First, to find \( f'(x) \), then repeat the process on \( f'(x) \) to determine \( f''(x) \), and so forth, until you reach the required third derivative, \( f'''(x) \). Higher-order derivatives help solve complex real-world problems such as analyzing the stability of structures or determining fluid dynamics.
Trigonometric Functions
Trigonometric functions like sine, cosine, and tangent are fundamental in calculus and have unique derivatives.
These derivatives help in both theoretical mathematics and practical applications, like physics and engineering.
Here are the primary derivatives you should be familiar with:
  • The derivative of \( \sin(x) \) is \( \cos(x) \).
  • The derivative of \( \cos(x) \) is \( -\sin(x) \).
  • The derivative of \( \tan(x) \) is \( \sec^2(x) \).
Understanding these derivatives is vital for complex problems in wave motion, oscillations, and alternating current circuits.
Another important point is using trigonometric identities when finding derivatives because they can simplify expressions before differentiating.
Keeping these derivatives fresh in your mind will make more complicated problems feel much simpler, especially as you progress in calculus.
Hyperbolic Functions
Hyperbolic functions, such as \( \sinh(x) \) and \( \cosh(x) \), have properties similar to trigonometric functions.
They often appear in calculus, especially in dealing with exponential growth, area hyperbolas, or in the field of engineering.
Here's a brief overview:
  • The derivative of \( \sinh(x) \) is \( \cosh(x) \).
  • The derivative of \( \cosh(x) \) is \( \sinh(x) \).
  • The derivative of \( \tanh(x) \) is \( \text{sech}^2(x) \).
Just like their counterparts in trigonometry, hyperbolic functions have identities,
making it easier to manipulate and simplify expressions involving them.
These derivatives are significant in various scenarios, such as modeling the shapes of cables in suspension bridges and in solving certain differential equations. Recognizing the parallels and differences between trigonometric and hyperbolic functions can greatly enhance your calculus toolkit.

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Most popular questions from this chapter

For each of the following functions \(f\), show that \(f^{-1}\) is differentiable and determine its derivative: (a) \(f(x)=\cos x, x \in(0, \pi)\); (b) \(f(x)=\sinh x, x \in \mathbb{R}\). Most equations of the form \(y=f(x)\) cannot be solved explicitly to give \(x\) as some formula involving \(y\) alone. The Inverse Function Rule is often used in such situations to solve problems that would otherwise be intractable. The following problem illustrates this type of application.

Determine whether the function $$ f(x)= \begin{cases}-x^{2}, & -2 \leq x<0 \\ x^{4}, & 0 \leq x<1 \\ x^{3}, & 1 \leq x \leq 2 \\ 0, & x>2\end{cases} $$ is differentiable at the points \(c=-2,0,1\) and 2, and determine the corresponding derivatives when they exist. Hint: \(\quad\) Sketch the graph \(y=f(x)\) first. Remarks Just as with continuity, the definition of differentiability involves a function These remarks are analogous \(f\) defined on a set in \(\mathbb{R}\), the domain \(A\) (say), that maps \(A\) to another set in \(\mathbb{R}\), to similar remarks for the codomain \(B\) (say). Sub-section 4.1.1. 1\. Let \(f\) and \(g\) be functions defined on open intervals \(I\) and \(J\), respectively, where \(I \supseteq J ;\) and let \(f(x)=g(x)\) on \(J .\) Technically \(g\) is a different function from \(f\). However, if \(f\) is differentiable at an interior point \(c\) of \(J\), it is a simple matter of some definition checking to verify that \(g\) too is differentiable at \(c .\) Similarly, if \(f\) is non- differentiable at \(c, g\) too is nondifferentiable at \(c\). 2\. The underlying point here is that differentiability at a point is a local property. It is only the behaviour of the function near that point that determines whether it is differentiable at the point.

Prove that the following functions \(f\) are not differentiable at the given point \(c\) : (a) \(f(x)=|x|^{\frac{1}{2}}, x \in \mathbb{R}, c=0\); (b) \(f(x)=[x], x \in \mathbb{R}, c=1\). Here \([x]\) denotes the integer Looking back at Example 2 , it looks as though chords joining the origin to points \((h, f(h))\) have slopes that tend to a limit 1 if \(h \rightarrow 0^{+}\), whereas they have slopes that tend to a limit \(-1\) if \(h \rightarrow 0^{-}\). This suggests the concept of one-sided derivatives that will be useful in our work later on. Definitions Let \(f\) be defined on an interval \(I\), and \(c \in I .\) Then the left derivative of \(f\) at \(c\) is $$ f_{L}^{\prime}(c)=\lim _{x \rightarrow c^{-}} \frac{f(x)-f(c)}{x-c} \text { All these definitions are } $$ analogous to definitions for continuity that you met in provided that this limit exists. In this case, we say that \(f\) is differentiable on the left at \(c\). Similarly, the right derivative of \(f\) at \(c\) is $$ f_{R}^{\prime}(c)=\lim _{x \rightarrow c^{+}} \frac{f(x)-f(c)}{x-c} \text { or } f_{R}^{\prime}(c)=\lim _{h \rightarrow 0^{+}} Q(h), $$ provided that this limit exists. In this case, we say that \(f\) is differentiable on the right at \(c\). A function \(f\) whose domain contains an interval \(I\) is differentiable on \(I\) if it is differentiable at each interior point of \(I\), differentiable on the right at the left end-point of \(I\) (if this belongs to \(I\) ) and differentiable on the left at the right end-point of \(I\) (if this belongs to \(I\) ). | {:[," Theorem 1 "],[,c" as an interior point is differentiable at "c" if and only if "f" is both differenti- "],[," able on the left at "c" and differentiable on the right at "c" AND "],[,f_(L)^(')(c),=f_(R)^(')(c).]:} | We omit a proof of this straight-forward result. The common value is simply f^(')(c). | | :--- | :--- | $$ \begin{aligned} &=\frac{\left\\{(-1+h)+(-1+h)^{2}\right\\}-\left\\{(-1)+(-1)^{2}\right\\}}{h} \\\ &=\frac{-h+h^{2}}{h} \\ &=-1+h \rightarrow-1 \quad \text { as } h \rightarrow 0^{+} \end{aligned} $$ It follows that \(f\) is differentiable on the right at \(-1\), and \(f_{R}^{\prime}(-1)=-1\) At 0, the function is defined on either side of 0 , but by a different formula; we therefore examine each side separately. At 0, for \(0

Find the derivative of each of the following functions: (a) \(f(x)=x^{7}-2 x^{4}+3 x^{3}-5 x+1, \quad x \in \mathbb{R}\); (b) \(f(x)=\frac{x^{2}+1}{x^{3}-1}, \quad x \in \mathbb{R}-\\{1\\}\); (c) \(f(x)=2 \sin x \cos x, \quad x \in \mathbb{R} ;\) (d) \(f(x)=\frac{e^{x}}{3+\sin x-2 \cos x}, \quad x \in \mathbb{R}\).

Find the derivative of each of the following functions: (a) \(f(x)=\tan x, \quad x \in \mathbb{R}-\left\\{\pm \frac{1}{2} \pi, \pm \frac{3}{2} \pi, \pm \frac{5}{2} \pi, \ldots\right\\}\) (b) \(f(x)=\operatorname{cosec} x, \quad x \in \mathbb{R}-\\{0, \pm \pi, \pm 2 \pi, \ldots\\}\) (c) \(f(x)=\sec x, \quad x \in \mathbb{R}-\left\\{\pm \frac{1}{2} \pi, \pm \frac{3}{2} \pi, \pm \frac{5}{2} \pi, \ldots\right\\}\) (d) \(f(x)=\cot x, \quad x \in \mathbb{R}-\\{0, \pm \pi, \pm 2 \pi, \ldots\\} .\)
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