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\(3-26=\) Differentiate. $$f(x)=\sin x+\frac{1}{2} \cot x$$

Short Answer

Expert verified
The derivative is \( f'(x) = \cos x - \frac{1}{2} \csc^2 x \).

Step by step solution

01

Differentiate the sine function

The derivative of \( \sin x \) is \( \cos x \). Thus, \( \frac{d}{dx} \sin x = \cos x \).
02

Differentiate the cotangent function

The derivative of \( \cot x \) is \( -\csc^2 x \). Therefore, given the function \( \frac{1}{2} \cot x \), its derivative is \( \frac{1}{2} \times -\csc^2 x = -\frac{1}{2} \csc^2 x \).
03

Combine derivatives

Now combine the derivatives of both terms to differentiate the entire function: \( f'(x) = \cos x - \frac{1}{2} \csc^2 x \).

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Key Concepts

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

Trigonometric Derivatives
Trigonometric derivatives are essential when working with calculus problems involving trigonometric functions. These derivatives allow us to find the rate of change of trigonometric functions with respect to a variable, typically denoted as \( x \). Understanding how each trigonometric function behaves when differentiated is crucial for successful problem-solving.

Here's a quick summary of a few common trigonometric derivatives:
  • The derivative of \( \sin x \) is \( \cos x \).
  • The derivative of \( \cos x \) is \( -\sin x \).
  • The derivative of \( \tan x \) is \( \sec^2 x \).
  • Finally, the derivative of \( \cot x \) is \( -\csc^2 x \).
Whenever you see a trigonometric function in calculus, recognizing these derivatives is crucial to solving the problem accurately. Practicing these ensures that you're ready to tackle any calculus challenges involving trigonometry.
Sine Function
The sine function, represented mathematically as \( \sin x \), is one of the fundamental trigonometric functions. When differentiating \( \sin x \), you obtain \( \cos x \). This means that the rate at which \( \sin x \) changes with respect to \( x \) is the same as \( \cos x \).

The sine function is periodic with a period of \( 2\pi \), and it oscillates between -1 and 1. Understanding the properties and behavior of the sine function is helpful when you encounter these expressions in differentiation.
  • Periodicity: The function completes a full cycle every \( 2\pi \) units.
  • Amplitude: The maximum and minimum values are 1 and -1 respectively.
  • Symmetry: The sine function is an odd function, meaning \( \sin(-x) = -\sin x \).
These essential features of the sine function can greatly simplify many calculus problems, especially those involving derivatives.
Cotangent Function
The cotangent function, or \( \cot x \), is another key trigonometric function. It is the reciprocal of the tangent function and can be defined as \( \cot x = \frac{1}{\tan x} = \frac{\cos x}{\sin x} \).

When differentiating \( \cot x \), the result is \( -\csc^2 x \). Here are a few helpful aspects to remember about the cotangent function:
  • **Reciprocal Function:** Since \( \cot x \) is \( 1/\tan x \), its behavior is related to the tangent function.
  • **Periodicity:** Like tangent and sine, \( \cot x \) is periodic, but with a period of \( \pi \).
  • **Symmetry:** It is an odd function, where \( \cot(-x) = -\cot x \).
These properties make \( \cot x \) a versatile function that appears in various trigonometric calculus problems. Understanding and memorizing its derivative and properties will make solving these problems more intuitive.
Chain Rule in Differentiation
The chain rule is an essential rule in calculus used to differentiate compositions of functions. While the provided exercise doesn't involve this principle directly, an understanding of the chain rule is crucial for complex calculus problems.

The chain rule states that if a function \( y \) is composed of another function \( u \), such that \( y = f(u) \) and \( u = g(x) \), then the derivative of \( y \) with respect to \( x \) is given by the product of their derivatives: \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \).For practical scenarios, if you have a function like \( \sin(g(x)) \), apply the chain rule as follows:
  • First, find the derivative of the outer function. Here, \( \sin \), which is \( \cos(g(x)) \).
  • Then multiply by the derivative of the inner function \( g(x) \).
The chain rule is an essential tool for differentiation in calculus, helping us to manage compositions of functions and ensure that we can simplify complex derivatives efficiently.

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