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In Exercises \(1-18,\) find \(d y / d x\) $$ y=\frac{4}{\cos x}+\frac{1}{\tan x} $$

Short Answer

Expert verified
\( \frac{dy}{dx} = 4 \sec x \tan x - \csc^2 x \).

Step by step solution

01

Identify the Differentiation Rule

The function given is a sum of two terms involving trigonometric functions. We'll need to use the differentiation rules for trigonometric functions, specifically the derivatives of \(\cos x\) and \(\tan x\). The relevant derivatives are: \((\cos x)' = -\sin x\) and \((\tan x)' = \sec^2 x\).
02

Differentiate Each Term Separately

Let's differentiate each term separately:1. The first term is \(\frac{4}{\cos x} = 4 \cdot \sec x\). The derivative of \(\sec x\) is \(\sec x \cdot \tan x\), so the derivative of \(4 \cdot \sec x\) is \(4 \cdot \sec x \cdot \tan x\).2. The second term is \(\frac{1}{\tan x} = \cot x\). The derivative of \(\cot x\) is \(-\csc^2 x\). Therefore, the derivative of \(\cot x\) is \(-\csc^2 x\).
03

Combine the Derivatives

Combine the derivatives of both terms:\[ \frac{d}{dx} \left( \frac{4}{\cos x} + \frac{1}{\tan x} \right) = \frac{d}{dx}(4 \cdot \sec x) + \frac{d}{dx}(\cot x) \]This will result in:\[ 4 \cdot \sec x \cdot \tan x - \csc^2 x \].
04

Write the Final Result

The derivative \( \frac{dy}{dx} \) of the function \( y \) is:\[ \frac{dy}{dx} = 4 \cdot \sec x \cdot \tan x - \csc^2 x \].

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Key Concepts

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

Trigonometric Functions
Trigonometric functions are fundamental in mathematics, especially in calculus. They express relationships between the angles and sides of triangles. The most common trigonometric functions are sine (\(\sin x\)), cosine (\(\cos x\)), and tangent (\(\tan x\)). These are critical to understanding wave patterns, circular movements, and oscillations. Other related functions include secant (\(\sec x = \frac{1}{\cos x}\)), cosecant (\(\csc x = \frac{1}{\sin x}\)), and cotangent (\(\cot x = \frac{1}{\tan x}\)).
These functions are pivotal when dealing with trigonometry in calculus since they are used in various derivative and integration problems. It's important to remember their basic identities and rules, as they become the building blocks for more complex operations.
  • The sine function (\(\sin x\)) is an odd function with a period of \(2\pi\). It's often used in calculating heights or radii.
  • The cosine function (\(\cos x\)) is an even function and also has a period of \(2\pi\). It's typically used to find distances along axes.
  • The tangent function (\(\tan x = \frac{\sin x}{\cos x}\)) represents the slope of the line related to the angle, useful in many practical situations.
Understanding these basics makes it easier to tackle any problem involving trigonometric functions, whether simple or complex.
Differentiation Technique
Differentiation is a crucial technique used in calculus to determine how a function changes as its input changes. When dealing with trigonometric functions, it's essential to know specific derivatives. These derivatives tell us the rate of change of these functions:
  • The derivative of sine, \(\sin x\), is \(\cos x\).
  • The derivative of cosine, \(\cos x\), is \(-\sin x\).
  • For tangent, \(\tan x\), the derivative is \(\sec^2 x\).
  • The derivative of secant, \(\sec x\), is \(\sec x \cdot \tan x\).
  • The derivative of cotangent, \(\cot x\), is \(-\csc^2 x\).
  • Lastly, the derivative of cosecant, \(\csc x\), is \(-\csc x \cdot \cot x\).
These derivative rules are derived from the fundamental definitions and properties of the trigonometric functions. Knowing these allows us to differentiate complex expressions that include combinations of trigonometric functions.
In calculus, understanding and applying these rules is necessary to solve real-world problems involving rates of change, such as motion dynamics or signal processing.
Calculus Exercise
Calculus exercises, like the one provided, often require applying a combination of algebraic manipulation and differentiation rules. In this instance, the exercise involved finding the derivative of a function composed of trigonometric terms:
  • The first term, \(\frac{4}{\cos x}\), can be rewritten using secant as \(4 \cdot \sec x\). The derivative for \(\sec x\) helps us here: \(\sec x \cdot \tan x\)
  • The second term, \(\frac{1}{\tan x}\), simplifies to \(\cot x\), and we use the derivative of \(\cot x\): \(-\csc^2 x\)
Applying these derivatives separately to each term simplifies the process and makes it more manageable. Finally, combining results gives us the overall derivative of the function in question:\[\frac{dy}{dx} = 4 \cdot \sec x \cdot \tan x - \csc^2 x\]Practicing these types of problems enhances your understanding and proficiency in calculus. The key is breaking down the function into parts you can manage individually and then synthesizing that information into the final derivative.

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Most popular questions from this chapter

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Diagonals If \(x, y,\) and \(z\) are lengths of the edges of a rectangular box, the common length of the box's diagonals is \(s=\) \(\sqrt{x^{2}+y^{2}+z^{2}}\) a. Assuming that \(x, y,\) and \(z\) are differentiable functions of \(t\) how is \(d s / d t\) related to \(d x / d t, d y / d t,\) and \(d z / d t\) ? b. How is \(d s / d t\) related to \(d y / d t\) and \(d z / d t\) if \(x\) is constant? c. How are \(d x / d t, d y / d t,\) and \(d z / d t\) related if \(s\) is constant?

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