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Find \(f^{\prime}(x)\). $$f(x)=\sec x \tan x$$

Short Answer

Expert verified
The derivative is \( f'(x) = \sec x \tan^2 x + \sec^3 x \).

Step by step solution

01

Recall the Product Rule

To find the derivative of a product of two functions, remember the product rule: \[(uv)' = u'v + uv'\]Here, we will treat \( u = \sec x \) and \( v = \tan x \).
02

Differentiate \( u = \sec x \)

The derivative of \( \sec x \) is \( \sec x \tan x \). Hence, \[u' = \sec x \tan x\].
03

Differentiate \( v = \tan x \)

The derivative of \( \tan x \) is \( \sec^2 x \). Hence, \[v' = \sec^2 x\].
04

Substitute into the Product Rule Formula

Substitute \( u \), \( v \), \( u' \), and \( v' \) into the product rule formula:\[f'(x) = (\sec x \tan x)(\tan x) + (\sec x)(\sec^2 x)\].
05

Simplify the Expression

Simplify the expression by combining terms:- The first term becomes \( \sec x \tan^2 x \).- The second term becomes \( \sec^3 x \).So, the derivative of \( f(x) = \sec x \tan x \) is:\[f'(x) = \sec x \tan^2 x + \sec^3 x\].

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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 we have a function that is the product of two smaller functions, we use the product rule to differentiate it. The product rule is important because it lets us find derivatives of more complex functions easily. The rule states: \((uv)' = u'v + uv'\). This means that the derivative of two functions multiplied together \(u\) and \(v\) is \(u'v\) plus \(uv'\). Here is how it works:
  • \(u\): One of the functions you are multiplying, such as \(\sec x\) in our case.
  • \(v\): The other function you are multiplying, which for our problem is \(\tan x\).
  • \(u'\): The derivative of \(u\), or \(\sec x\tan x\) in this exercise.
  • \(v'\): The derivative of \(v\), or \(\sec^2 x\) here.
This step-by-step approach given by the product rule is a systematic way to handle products of functions while differentiating. It's one of the core tools in calculus, especially when dealing with intertwined functions. Break complex expressions down into products, and apply this golden rule. This allows for easier calculation and understanding.
Why Trigonometric Functions Matter
Trigonometric functions, like \(\sec x\) and \(\tan x\), appear frequently in calculus due to their unique properties and periodic nature. They are the backbone of many calculus problems both simple and complex.
  • \(\sec x\): Also known as the secant function, is defined as \(1/\cos x\). It's useful because it naturally complements the cosine function and appears as a derivative in calculus problems.
  • \(\tan x\): The tangent function is the ratio of sine to cosine, \(\sin x / \cos x\). This function offers a different perspective on angle measurements and slopes, crucial for calculus.
In the context of derivatives:
  • Derivative of \(\sec x\): The derivative is \(\sec x \tan x\). This result comes from the product of secant and tangent, reflecting their inherent relationship.
  • Derivative of \(\tan x\): This is \(\sec^2 x\). An important derivative that helps describe how the tangent function changes.
Understanding these trigonometric relationships and derivatives is fundamental. They help simplify complex function differentiation, which is essential in many fields, such as physics and engineering.
Differentiation Steps Made Simple
Differentiation is about finding how a function changes at any point. The approach usually involves systematic steps that you can follow to simplify the process.

Step-by-step Differentiation

Using the product rule along with the knowledge of derivatives, you can easily tackle derivative problems involving trigonometric functions. Here's a breakdown:
  • Identify the Functions: Recognize that your function \(f(x) = \sec x \tan x\) is the product of \(u = \sec x\) and \(v = \tan x\).
  • Find Derivatives of Each Part: Compute \(u' = \sec x \tan x\) and \(v' = \sec^2 x\).
  • Apply the Product Rule: Substitute these into the product rule to get \(f'(x) = (\sec x \tan x)(\tan x) + (\sec x)(\sec^2 x)\).
  • Simplify: Combine and simplify the terms to get \(f'(x) = \sec x \tan^2 x + \sec^3 x\).
These steps highlight the clarity of following a structured method for differentiation. This makes solving even complex problems approachable and manageable. With practice, these steps become intuitive, allowing you to swiftly identify the necessary actions for any given function.

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Most popular questions from this chapter

(a) Use a graphing utility to obtain the graph of the function \(f(x)=\sin x^{2} \cos x\) over the interval \([-\pi / 2, \pi / 2]\) (b) Use the graph in part (a) to make a rough sketch of the graph of \(f^{\prime}\) over the interval. (c) Find \(f^{\prime}(x)\), and then check your work in part (b) by using the graphing utility to obtain the graph of \(f^{\prime}\) over the interval. (d) Find the equation of the tangent line to the graph of \(f\) at \(x=1,\) and graph \(f\) and the tangent line together over the interval.

You are asked in these exercises to determine whether a piecewise-defined function \(f\) is differentiable at a value \(x=x_{0}\) where \(f\) is defined by different formulas on different sides of \(x_{0}\). You may use without proof the following result, which is a consequence of the Mean-Value Theorem (discussed in Section 4.8 ). Theorem. Let \(f\) be continuous at \(x_{0}\) and suppose that \(\lim _{x \rightarrow x_{0}} f^{\prime}(x)\) exists. Then \(f\) is differentiable at \(x_{0}\) and \(f^{\prime}\left(x_{0}\right)=\lim _{x \rightarrow \infty} f^{\prime}(x)\) Let $$ f(x)=\left\\{\begin{array}{ll} x^{2}, & x \leq 1 \\ \sqrt{x}, & x>1 \end{array}\right. $$ Determine whether \(f\) is differentiable at \(x=1 .\) If so, find the value of the derivative there.

Find \(f^{\prime}(x)\) $$f(x)=\cos ^{2}(3 \sqrt{x})$$

An object is released from rest (its initial velocity is zero) from the Empire State Building at a height of \(1250 \mathrm{ft}\) above street level (Figure Ex-26 on the next page). The height of the object can be modeled by the position function \(s=f(t)=1250-16 t^{2}.\) (a) Verify that the object is still falling at \(t=5 \mathrm{s}.\) (b) Find the average velocity of the object over the time interval from \(t=5\) to \(t=6 \mathrm{s}.\) (c) Find the object's instantaneous velocity at time \(t=5 \mathrm{s}.\) (Check your book to see figure)

Find \(d y / d x\) $$y=\left(\frac{x-5}{2 x+1}\right)^{3}$$

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