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Chapter 3: Problem 10

Find \(\frac{d^{2}}{d x^{2}}(\sec x)\).

Short Answer

Expert verified
Question: Determine the second derivative of the function \(\sec x\). Answer: The second derivative of the given function is \(\frac{d^2}{dx^2}(\sec x) = \sec x (\tan^2 x + \sec^2 x)\).

Step by step solution

01

Find the first derivative of \(\sec x\)

The first step is to find the derivative of \(\sec x\) with respect to \(x\), denoted as \(\frac{d}{dx}(\sec x)\). Recall that the derivative of \(\sec x\) is given by: $$\frac{d}{dx}(\sec x) = \sec x \tan x$$
02

Find the second derivative of \(\sec x\)

Now we need to find the second derivative of the function, denoted as \(\frac{d^2}{dx^2}(\sec x)\), by differentiating the first derivative, \(\sec x \tan x\). To do this, we use the product rule, given by: $$\frac{d}{dx}(uv) = u'v + uv'$$ where \(u = \sec x\) and \(v = \tan x\). We already know the derivative of \(\sec x\) but we need to know the derivative of \(\tan x\). Recall that the derivative of \(\tan x\) is given by: $$\frac{d}{dx}(\tan x) = \sec^2 x$$ Now apply the product rule for the second derivative: $$\frac{d^2}{dx^2}(\sec x) = \frac{d}{dx}(\sec x \tan x) = (\sec x)' \tan x + \sec x (\tan x)' = \sec x \tan x \cdot \tan x + \sec x \cdot \sec^2 x$$ Simplify the expression: $$\frac{d^2}{dx^2}(\sec x) = \sec x \tan^2 x + \sec^3 x$$ The second derivative of the given function is: $$\frac{d^2}{dx^2}(\sec x) = \sec x (\tan^2 x + \sec^2 x)$$

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Key Concepts

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

Product Rule
Understanding the product rule is critical when differentiating products of two functions, which is a common scenario in calculus. When you have a function that is the product of two other functions, say f(x) = u(x) * v(x), the product rule states that the derivative of f(x) with respect to x is f'(x) = u' * v + u * v'. In other words, you find the derivative of each function separately and then combine them, one multiplied by the derivative of the other.

This rule becomes even more important when finding higher order derivatives, such as the second derivative. In these cases, you'll apply the product rule to your result from the first derivative, which often still contains products of functions. Memorize the product rule as it is a fundamental tool in the differentiation toolkit.
First Derivative
The first derivative of a function is essentially the heart of differential calculus. It represents the rate at which a function is changing at any given point. When you have the function y = sec(x), the first derivative, denoted as \(\frac{d}{dx}(\sec x)\) or \(\sec'(x)\), indicates how the secant (sec) of angle x changes with a small change in x. For trigonometric functions like the secant function, finding the first derivative often requires you to remember specific formulas. In this case, the first derivative of \(\sec x\) is \(\sec x \tan x\), which sets the stage for finding higher order derivatives by applying additional differentiation techniques.
Trigonometric Derivatives
When dealing with trigonometric functions such as sine, cosine, secant, and tangent, their derivatives are not as straightforward as polynomials. Special formulas have been established for these derivatives, which are essential to remember. For example:
  • The derivative of \(\sin x\) is \(\cos x\)
  • The derivative of \(\cos x\) is \( -\sin x\)
  • The derivative of \(\tan x\) is \(\sec^2 x\)
  • The derivative of \(\sec x\) is \(\sec x \tan x\).
Being familiar with these formulas allows you to differentiate trigonometric functions easily. This becomes especially useful when these functions are part of more complex expressions or when you are calculating higher derivatives, as seen in our exercise with the second derivative of \(\sec x\).
Differentiation Techniques
Differentiation is not a one-size-fits-all operation, and thus, various techniques need to be applied depending on the function at hand. Besides the product rule, there are other techniques such as:
  • Chain rule: Used when differentiating compositions of functions.
  • Quotient rule: Similar to the product rule, but used for quotients of functions.
  • Implicit differentiation: Used when a function is not explicitly solved for one variable.
  • Higher-order derivatives: Finding the second derivative or beyond often requires reapplying the aforementioned techniques or recognizing patterns in repetitive differentiation.
To effectively differentiate functions, one needs to identify which technique or combination of techniques apply to a given problem. Mastery of these techniques will greatly assist in tackling more challenging calculus problems, such as finding the second derivative of a function.

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Most popular questions from this chapter

The following equations implicitly define one or more functions. a. Find \(\frac{d y}{d x}\) using implicit differentiation. b. Solve the given equation for \(y\) to identify the implicitly defined functions \(y=f_{1}(x), y=f_{2}(x), \ldots.\) c. Use the functions found in part (b) to graph the given equation. \(x+y^{3}-x y=1\) (Hint: Rewrite as \(y^{3}-1=x y-x\) and then factor both sides.)

The hands of the clock in the tower of the Houses of Parliament in London are approximately \(3 \mathrm{m}\) and \(2.5 \mathrm{m}\) in length. How fast is the distance between the tips of the hands changing at \(9.00 ?\) (Hint: Use the Law of Cosines.)

The population of a culture of cells after \(t\) days is approximated by the function \(P(t)=\frac{1600}{1+7 e^{-0.02 t}},\) for \(t \geq 0\) a. Graph the population function. b. What is the average growth rate during the first 10 days? c. Looking at the graph, when does the growth rate appear to be a maximum? d. Differentiate the population function to determine the growth rate function \(P^{\prime}(t)\) e. Graph the growth rate. When is it a maximum and what is the population at the time that the growth rate is a maximum?

Suppose \(f\) is differentiable on [-2,2] with \(f^{\prime}(0)=3\) and \(f^{\prime}(1)=5 .\) Let \(g(x)=f(\sin x)\) Evaluate the following expressions. a. \(g^{\prime}(0)\) b. \(g^{\prime}\left(\frac{\pi}{2}\right)\) c. \(g^{\prime}(\pi)\)

The following equations implicitly define one or more functions. a. Find \(\frac{d y}{d x}\) using implicit differentiation. b. Solve the given equation for \(y\) to identify the implicitly defined functions \(y=f_{1}(x), y=f_{2}(x), \ldots.\) c. Use the functions found in part (b) to graph the given equation. \(x^{4}=2\left(x^{2}-y^{2}\right)\) (eight curve)
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