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Chapter 2: Problem 26

Find the derivative of the function. $$ g(x)=e^{-2 x} \cos 3 x $$

Short Answer

Expert verified
The derivative of the function \(g(x) = e^{-2x} \cos 3x\) is \(g'(x) = -2e^{-2x}\cos{3x} - 3e^{-2x}\sin{3x}\).

Step by step solution

01

Apply the product rule

Let's denote \(u(x) = e^{-2x}\) and \(v(x) = \cos{3x}\). We can now rewrite our function as \(g(x) = u(x) \cdot v(x)\). To find the derivative of the function \(g(x)\), we will apply the product rule: \[ \frac{d}{dx}(u(x) \cdot v(x)) = u'(x) \cdot v(x) + u(x) \cdot v'(x) \]
02

Find the derivatives of u(x) and v(x)

We will now find the derivatives of the functions u(x) and v(x). For \(u(x) = e^{-2x}\), apply the chain rule: \[ u'(x) = \frac{d}{dx}(e^{-2x}) = -2e^{-2x} \] For \(v(x) = \cos{3x}\), apply the chain rule: \[ v'(x) = \frac{d}{dx}(\cos{3x}) = -3\sin{3x} \]
03

Use the product rule with the derivatives

Now we will substitute the derivatives u'(x) and v'(x) into the product rule formula: \[ g'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) = (-2e^{-2x})(\cos{3x}) + (e^{-2x})(-3\sin{3x}) \]
04

Simplify the expression

Now, let's simplify the expression: \[ g'(x) = -2e^{-2x}\cos{3x} - 3e^{-2x}\sin{3x} \] So, the derivative of the function \(g(x) = e^{-2 x} \cos 3 x\) is: \[ g'(x) = -2e^{-2x}\cos{3x} - 3e^{-2x}\sin{3x} \]

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Key Concepts

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

Understanding the Product Rule
When you need to find the derivative of a function that is the product of two other functions, the product rule comes into play. This rule states that if you have a function \(g(x) = u(x) \cdot v(x)\), then the derivative \(g'(x)\) is computed as:
  • First derivative of \(u(x)\), multiplied by \(v(x)\)
  • Plus \(u(x)\), multiplied by the derivative of \(v(x)\)
Mathematically, it's written as:\[\frac{d}{dx}(u(x) \cdot v(x)) = u'(x) \cdot v(x) + u(x) \cdot v'(x)\]This means you need to differentiate each function separately and then combine them using this formula. It鈥檚 an essential rule for tackling complex functions, especially when dealing with combinations like trigonometric and exponential functions.
Making Sense of the Chain Rule
The chain rule is used when you have a function within a function鈥攃ommonly called a composition of functions. If you have a function \(f(g(x))\), the chain rule helps you find its derivative. The rule tells us to differentiate the outer function first and then multiply by the derivative of the inner function.
  • Differentiate the outer function while keeping the inner function unchanged.
  • Multiply by the derivative of the inner function.
This technique is simple but very powerful. For example, with \(u(x) = e^{-2x}\), the outer function is the exponential function \(e^u\), and the inner function is \(-2x\). Differentiating gives \(-2e^{-2x}\) because you multiply the derivative of the outer by the derivative of the inner.
The Role of Trigonometric Functions
Trigonometric functions, like sine and cosine, have specific derivatives that are important in calculus. For a function like \(\cos{3x}\), the derivative involves the chain rule since it's a composition function. The derivative of \(\cos{u}\) is \(-\sin{u}\). Thus, finding \(v'(x)\) for \(v(x) = \cos{3x}\) means:
  • Firstly, differentiate \(\cos{u}\) to get \(-\sin{u}\).
  • Secondly, multiply by the derivative of the inner, which is the \(3x\) in this case, giving \(-3\sin{3x}\).
Understanding these derivatives helps you solve a wide range of problems and find the rate of change of trigonometric functions quickly.
Understanding Exponential Functions
Exponential functions often appear in calculus problems due to their constant rate of change characteristics. When differentiating an exponential function like \(e^{-2x}\), we utilize the chain rule.
  • The derivative of \(e^x\) is itself, \(e^x\).
  • For \(e^{-2x}\), apply the chain rule: the outer function is \(e^x\) and inner function is \(-2x\), thus the derivative becomes \(-2e^{-2x}\).
This concept is pivotal in both natural processes and mathematical modeling. Knowing how to differentiate exponential functions allows you to tackle problems in calculus with greater confidence.

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Most popular questions from this chapter

Let \(g\) denote the inverse of the function \(f\). (a) Show that the point \((a, b)\) lies on the graph of \(f .\) (b) Find \(g^{\prime}(b)\) $$ f(x)=2 x+1 ; \quad(2,5) $$

Find an equation of the tangent line to the given curve at the indicated point. $$ x^{2} y^{2}=(y+1)^{2}\left(4-y^{2}\right) ; \quad(-2 \sqrt{3}, 1) $$

In Exercises, (a) find the equations of the tangent and the normal lines to the curve at the indicated point. (The normal line at a point on the curve is the line perpendicular to the tangent line at that point.) (b) Then use a graphing utility to plot the curve and the tangent and normal lines on the same screen. $$ x^{2}+y^{2}=9 ; \quad(-1,2 \sqrt{2}) $$

Find \(f^{\prime \prime}(x)\) if $$ f(x)=\left\\{\begin{array}{ll} x^{2} \sin \frac{1}{x} & \text { if } x \neq 0 \\ 0 & \text { if } x=0 \end{array}\right. $$ Does \(f^{\prime \prime}(0)\) exist?

Simple Harmonic Motion The equation of motion of a body executing simple harmonic motion is given by $$ x(t)=A \sin (\omega t+\phi) $$ where \(x\) (in feet) is the displacement of the body, \(A\) is the amplitude, \(\omega=\sqrt{k / m}, k\) is a constant, and \(m\) (in slugs) is the mass of the body. Find expressions for the velocity and acceleration of the body at time \(t\).
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