/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 Find the derivatives of the give... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the derivatives of the given functions. $$y=3 \cot 6 x$$

Short Answer

Expert verified
The derivative of \( y = 3 \cot(6x) \) is \( y' = -18 \csc^2(6x) \).

Step by step solution

01

Identify the Derivative Formula for Cotangent

To find the derivative of a cotangent function, recall that the derivative of \(\cot(x)\) is \(-\csc^2(x)\). This formula will be useful to differentiate \(\cot(6x)\).
02

Apply the Chain Rule

The function we are dealing with is \(y = 3 \cot(6x)\). Since \(\cot(6x)\) is a composite function involving \(6x\), we'll use the chain rule, which states that \((f(g(x)))' = f'(g(x)) \cdot g'(x)\).
03

Differentiate the Outer Function

First, differentiate the outer part: the derivative of \(3 \cot(u)\) with respect to \(u\) is \(-3 \csc^2(u)\). Here \(u = 6x\).
04

Differentiate the Inner Function

Next, differentiate the inner function \(u = 6x\) with respect to \(x\), which is simply \(6\).
05

Combine using the Chain Rule

Combine the derivatives from Step 3 and Step 4 using the chain rule: \(\frac{d}{dx} (3 \cot(6x)) = -3 \csc^2(6x) \cdot 6 = -18 \csc^2(6x)\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Derivatives
Derivatives are a central concept in calculus. By definition, the derivative represents how a function changes as its input changes, or more simply, the "rate of change". For example, if we have a function that describes the distance an object travels over time, the derivative of that function gives us the speed, or how fast the distance is changing with respect to time.
When finding derivatives, we often deal with standard functions, like linear or polynomial functions. But the challenge increases with more complex forms such as trigonometric functions, where we need specific rules.
  • Derivatives help us analyze various physical systems.
  • They’re used in a range of fields, from physics to economics.
Consider the cotangent function, as in this exercise. It isn't as straightforward as linear functions, and requires us to remember specific formulas and sometimes use rules like the chain rule.
Chain Rule
The chain rule is an essential tool in the differentiation of composite functions. A composite function is one where a function is applied to the result of another function, like in our exercise with \(\cot(6x)\). To differentiate these functions, we apply the chain rule, which states: \((f(g(x)))' = f'(g(x)) \cdot g'(x)\).
  • Think of the chain rule as peeling layers of an onion. You differentiate the outer layer first and then the inner.
  • This rule allows us to break down complex derivatives into simpler, manageable parts.
By applying the chain rule: we start by differentiating the outer function (\(\cot\)), followed by the inner function (\(6x\)). This results in a product of their derivatives, revealing the derivative of the entire composite function.
Trigonometric Functions
Trigonometric functions, including sine, cosine, and cotangent, play a vital role in calculus. Each has its own derivative, which is used frequently:
  • The derivative of \(\sin(x)\) is \(\cos(x)\).
  • The derivative of \(\cos(x)\) is \(-\sin(x)\).
  • The derivative of \(\cot(x)\) is \(-\csc^2(x)\).
These derivatives are foundational for understanding how trigonometric functions change. In practical terms, they allow us to model oscillations, waves, and other periodic phenomena. Memorizing these derivatives and knowing when to apply them is key to mastering calculus, especially when dealing with more complex trigonometric forms.
Cotangent Function
The cotangent function is a lesser-known trigonometric function compared to sine and cosine, but it’s equally important. Defined as \(\cot(x) = \frac{1}{\tan(x)} = \frac{\cos(x)}{\sin(x)}\), it expresses the ratio of the adjacent side to the opposite side in a right triangle. In calculus, the derivative of \(\cot(x)\) is essential for problems involving angular speed and periodic functions.
  • The derivative of the cotangent function is \(-\csc^2(x)\).
  • This means it describes changes in any particular cotangent curve as it varies with input \(x\).
The negative sign in its derivative signifies how \(\cot(x)\) decreases as \(x\) increases, when considering its principal values. Therefore, understanding \(\cot(x)\) and its derivative helps in solving complex calculus problems that involve periodic changes.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Use the following information. The hyperbolic sine of \(u\) is defined as \(\sinh u=\frac{1}{2}\left(e^{u}-e^{-u}\right)\) Figure 27.37 shows the graph of \(y=\sinh x\) The hyperbolic cosine of \(u\) is defined as \(\cosh u=\frac{1}{2}\left(e^{\mu}+\mathrm{e}^{-u}\right)\) Figure 27.38 shows the graph of \(y=\cosh x\) These functions are called hyperbolic functions since, if \(x=\cosh u\) and \(y=\sinh u, x\) and \(y\) satisfy the equation of the hyperbola \(x^{2}-y^{2}=1\) Show that \(\frac{d}{d x} \sinh u=\cosh u \frac{d u}{d x}\) and \(\frac{d}{d x} \cosh u=\sinh u \frac{d u}{d x}\) where \(u\) is a function of \(x\)

Solve the given problems. Use differentials to estimate the value of \(\sin 31^{\circ}\)

Solve the given problems. The number \(N\) of reflections of a light ray passing through an optic fiber of length \(L\) and diameter \(d\) is \(N=\frac{L \sin \theta}{d \sqrt{n^{2}-\sin ^{2} \theta}}\) Here, \(n\) is the index of refraction of the fiber, and \(\theta\) is the angle between the light ray and the fiber's axis. Find \(d N / d \theta\)

Solve the given problems. Find the slope of a line tangent to the curve of \(y=\frac{2 \sin 3 x}{x},\) where \(x=0.15 .\) Verify the result by using the derivative-evaluating feature of a calculator.

Solve the given problems by finding the appropriate derivative. A missile is launched and travels along a path that can be represented by \(y=\sqrt{x} .\) A radar tracking station is located \(2.00 \mathrm{km}\) directly behind the launch pad. Placing the launch pad at the origin and the radar station at \((-2.00,0),\) find the largest angle of elevation required of the radar to track the missile.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.