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Chapter 3: Problem 7

Find \(d y / d x\). $$f(x)=\sin x \tan x$$

Short Answer

Expert verified
\( \frac{dy}{dx} = \sin x (1 + \sec^2 x) \)

Step by step solution

01

Identify the function composition

The function given is \(f(x) = \sin x \tan x\), which is a product of two functions: \(u = \sin x\) and \(v = \tan x\). We will use the product rule to differentiate it.
02

Apply the Product Rule

The product rule states that if \(y = u \cdot v\), then the derivative \(\frac{dy}{dx} = u'v + uv'\). In this case, we need to find \(u'\) and \(v'\).
03

Differentiate \\(u = \sin x\\)

The derivative of \(u = \sin x\) is \(u' = \cos x\).
04

Differentiate \\(v = \tan x\\)

The derivative of \(v = \tan x\) is \(v' = \sec^2 x\).
05

Substitute in the Product Rule Formula

Substitute the derivatives into the product rule: \[\frac{dy}{dx} = (\cos x)(\tan x) + (\sin x)(\sec^2 x)\]
06

Simplify the Expression

Simplify the expression: \[\frac{dy}{dx} = \cos x \tan x + \sin x \sec^2 x\]Where \(\cos x \tan x = \sin x\), giving:\[\frac{dy}{dx} = \sin x + \sin x \sec^2 x\]Since \(\sec^2 x = 1 + \tan^2 x\), write as:\[\frac{dy}{dx} = \sin x (1 + \sec^2 x)\]

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Key Concepts

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

Differentiation
Differentiation is a fundamental concept in calculus. It involves finding the derivative of a function, which essentially measures how a function changes as its input changes.
This rate of change is crucial in understanding the behavior of various functions.
  • A derivative can tell you how steep a curve is at any given point.
  • Derivatives are used to find local maxima and minima, optimization problems, and to model real-world situations.
In the exercise, the goal is to find the derivative, or the rate of change, of a trigonometric function, specifically, \( f(x) = \sin x \tan x \).
Applying the product rule is necessary because this function is a product of two simpler trigonometric functions: \( \sin x \) and \( \tan x \).
The product rule for differentiation aids in calculating this derivative by breaking down the problem into manageable parts.
Trigonometric Derivatives
Trigonometric derivatives refer to the derivatives of trigonometric functions like sine, cosine, and tangent. These derivatives are pivotal for calculus problems involving periodic functions.
Here are the derivatives used in this exercise:
  • For \( u = \sin x \, \ u' = \cos x \)
  • For \( v = \tan x \, \ v' = \sec^2 x \)
Knowing these derivatives by heart can save you time during calculations. These functions' behavior often repeats due to their periodic nature.
It's useful to remember some key trigonometric identities as well, like \( \sec^2 x = 1 + \tan^2 x \).
These identities facilitate simplifying derivative expressions, providing essential insights into the dynamics of trigonometric functions.
Calculus
Calculus is the field of mathematics that studies change, and integration and differentiation are its two main branches.
Differentiation deals with rates of change and slopes of curves, whereas integration is about aggregation or accumulation.
  • Calculus allows us to understand and predict the behavior of complex systems in physics, engineering, biology, and economics.
  • In the given exercise, calculus concepts are applied to differentiate a product of trigonometric functions: \( \sin x \tan x \).
Beyond just finding derivatives, calculus provides tools for analyzing function behavior over an entire domain.
This can include identifying function continuity, points of inflection, and ascertaining increasing or decreasing intervals. Here, it helps simplify complex expressions, like transforming \( \sec^2 x \) into \( 1 + \tan^2 x \), which enhances understanding and problem-solving approaches in trigonometric contexts.

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Most popular questions from this chapter

Find \(d y / d t\) $$y=\cos \left(5 \sin \left(\frac{t}{3}\right)\right)$$

Find both \(d y / d x\) (treating \(y\) as a differentiable function of \(x\) ) and \(d x / d y\) (treating \(x\) as a differentiable function of \(y\) ). How do \(d y / d x\) and \(d x / d y\) seem to be related? Explain the relationship geometrically in terms of the graphs. $$x y^{3}+x^{2} y=6$$

Use a CAS to estimate the magnitude of the error in using the linearization in place of the function over a specified interval \(I\). Perform the following steps: a. Plot the function \(f\) over \(I\) b. Find the linearization \(L\) of the function at the point \(a\) c. Plot \(f\) and \(L\) together on a single graph. d. Plot the absolute error \(|f(x)-L(x)|\) over \(I\) and find its maximum value. e. From your graph in part (d), estimate as large a \(\delta>0\) as you can, satisfying \(|x-a|<\delta \quad \Rightarrow \quad|f(x)-L(x)|<\epsilon\) for \(\epsilon=0.5,0.1,\) and \(0.01 .\) Then check graphically to see if your \(\delta\) -estimate holds true. $$f(x)=\frac{x-1}{4 x^{2}+1}, \quad\left[-\frac{3}{4}, 1\right], \quad a=\frac{1}{2}$$

In the late 1860 s, Adolf Fick, a professor of physiology in the Faculty of Medicine in W眉rzberg, Germany, developed one of the methods we use today for measuring how much blood your heart pumps in a minute. Your cardiac output as you read this sentence is probably about \(7 \mathrm{L} / \mathrm{min} .\) At rest it is likely to be a bit under \(6 \mathrm{L} / \mathrm{min}\). If you are a trained marathon runner running a marathon, your cardiac output can be as high as \(30 \mathrm{L} / \mathrm{min}\). Your cardiac output can be calculated with the formula $$y=\frac{Q}{D},$$ where \(Q\) is the number of milliliters of \(\mathrm{CO}_{2}\) you exhale in a minute and \(D\) is the difference between the \(\mathrm{CO}_{2}\) concentration \((\mathrm{mL} / \mathrm{L})\) in the blood pumped to the lungs and the \(\mathrm{CO}_{2}\) concentration in the blood returning from the lungs. With \(Q=233 \mathrm{mL} / \mathrm{min}\) and \(D=97-56=41 \mathrm{mL} / \mathrm{L}\), \(y=\frac{233 \mathrm{mL} / \mathrm{min}}{41 \mathrm{mL} / \mathrm{L}} \approx 5.68 \mathrm{L} / \mathrm{min}\), fairly close to the \(6 \mathrm{L} / \mathrm{min}\) that most people have at basal (resting) conditions. (Data courtesy of J. Kenneth Herd, M.D., Quillan College of Medicine, East Tennessee State University.) Suppose that when \(Q=233\) and \(D=41,\) we also know that \(D\) is decreasing at the rate of 2 units a minute but that \(Q\) remains unchanged. What is happening to the cardiac output?

What happens if you can write a function as a composite in different ways? Do you get the same derivative each time? The Chain Rule says you should. Try it with the functions. Find \(d y / d x\) if \(y=x^{3 / 2}\) by using the Chain Rule with \(y\) as a composite of a. \(y=u^{3}\) and \(u=\sqrt{x}\) b. \(y=\sqrt{u}\) and \(u=x^{3}\)
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