/*! 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 45 In Problems \(1-58\), find the d... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In Problems \(1-58\), find the derivative with respect to the independent variable. $$ (x)=\frac{1}{\sin \left(3 x^{2}-1\right)} $$

Short Answer

Expert verified
The derivative is \( -6x \cdot \cot(3x^2 - 1) \cdot \csc(3x^2 - 1) \).

Step by step solution

01

Identify the Composite Function

The function given is \( f(x) = \frac{1}{\sin(3x^2 - 1)} \). This is a composite function, which means it is composed of an outer function \( g(u) = \frac{1}{u} \) and an inner function \( u(x) = \sin(3x^2 - 1) \).
02

Differentiate the Outer Function

First, gain the derivative of the outer function \( g(u) = \frac{1}{u} \) with respect to \( u \). The derivative is \( g'(u) = -\frac{1}{u^2} \).
03

Differentiate the Inner Function

Now, we need to differentiate the inner function \( u(x) = \sin(3x^2 - 1) \) with respect to \( x \). First, the derivative of \( \sin(v(x)) \) is \( \cos(v(x)) \cdot v'(x) \). Here, \( v(x) = 3x^2 - 1 \), so its derivative is \( v'(x) = 6x \). Thus, the derivative of \( u(x) \) is \( \cos(3x^2 - 1) \cdot 6x \).
04

Apply the Chain Rule

Using the chain rule to connect the derivatives from Steps 2 and 3. The chain rule states that \( f'(x) = g'(u(x)) \cdot u'(x) \). Therefore, \( f'(x) = \left(-\frac{1}{\sin^2(3x^2 - 1)}\right) \cdot \left(\cos(3x^2 - 1) \cdot 6x \right) \).
05

Simplify the Expression

Simplify the expression from Step 4: \( f'(x) = -6x \cdot \frac{\cos(3x^2 - 1)}{\sin^2(3x^2 - 1)} \). Recognizing that \( \frac{\cos(a)}{\sin^2(a)} = \cot(a) \cdot \csc(a) \), we can further write \( f'(x) = -6x \cdot \cot(3x^2 - 1) \cdot \csc(3x^2 - 1) \).

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.

Understanding the Chain Rule
The chain rule is a powerful technique when dealing with composite functions in calculus. When you need to differentiate a function that's been composed by layering more than one function together, the chain rule is your go-to method. Think of it like peeling layers off an onion; you handle one layer at a time.
For a composite function, let's say you have a function like \( f(g(x)) \). In this scenario, \( f \) is layered on top of \( g \), and both functions rely on the same variable, \( x \). The chain rule allows us to differentiate this layered or composite function efficiently by expressing the derivative as:
  • \( f'(g(x)) \cdot g'(x) \)
This process enables you to break down complex problems into simpler, more manageable parts so you can find their derivatives. Make sure you first find the derivative of the outer function and then follow up with the derivative of the inner function, always multiplying them together based on this rule.
Unpacking Composite Functions
Composite functions occur when one function is applied to the result of another function. For example, in the expression \( f(g(x)) \), \( g(x) \) is the inner function and \( f(u) \) (where \( u = g(x) \)) is the outer function. The expression is considered composite because you’re nesting one function inside another.
In situations such as the one provided, you are often working with functions within functions. Identifying the inner and outer components is crucial. To illustrate, our example function is \( f(x) = \frac{1}{\sin(3x^2 - 1)} \). Here:
  • Outer function: \( g(u) = \frac{1}{u} \)
  • Inner function: \( u(x) = \sin(3x^2 - 1) \)
Recognizing and breaking down these layers indicates which derivatives you'll need to compute separately before merging them using the chain rule.
Essential Differentiation Techniques
Differentiation techniques are foundational in calculus for finding the rates at which things change. Three primary techniques are most frequently used:
1. **The Power Rule**: Useful for power functions, states the derivative of \( x^n \) is \( nx^{n-1} \).
  • Simple yet effective when dealing directly with polynomial functions.
2. **The Product Rule**: Applied when differentiating products of two functions, stated as:
  • \((uv)' = u'v + uv'\)
This ensures each function's change is considered.
3. **The Chain Rule**: Discussed previously for composite functions, helps differentiate nested functions.
When approaching problems, select the proper rules based on the function's structure. In our specific example, employing the chain rule is critical as it facilitates breaking down and differentiating the composite function by its inner and outer parts. Differentiation techniques like these form the backbone of calculus analysis and problem-solving, enabling you to tackle even the most complex structures.

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

Graph each function and, on the basis of the graph, guess where the function is not differentiable. (Assume the largest possible domain.) $$ y=\frac{1}{2+x} $$

Assume that a species lives in a habitat that consists of many islands close to a mainland. The species occupies both the mainland and the islands, but, although it is present on the mainland at all times, it frequently goes extinct on the islands. Islands can be recolonized by migrants from the mainland. The following model keeps track of the fraction of islands occupied: Denote the fraction of islands occupied at time \(t\) by \(p(t)\). Assume that each island experiences a constant risk of extinction and that vacant islands (the fraction \(1-p\) ) are colonized from the mainland at a constant rate. Then $$ \frac{d p}{d t}=c(1-p)-e p $$ where \(c\) and \(e\) are positive constants. (a) The gain from colonization is \(f(p)=c(1-p)\) and the loss from extinction is \(g(p)=e p .\) Graph \(f(p)\) and \(g(p)\) for \(0 \leq p \leq 1\) in the same coordinate system. Explain why the two graphs intersect whenever \(e\) and \(c\) are both positive. Compute the point of intersection and interpret its biological meaning. (b) The parameter \(c\) measures how quickly a vacant island becomes colonized from the mainland. The closer the islands, the larger is the value of \(c\). Use your graph in (a) to explain what happens to the point of intersection of the two lines as \(c\) increases. Interpret your result in biological terms.

Approximate \(f(x)\) at a by the linear approximation $$L(x)=f(a)+f^{\prime}(a)(x-a)$$ $$ f(x)=\log x \text { at } a=1 $$

Graph each function and, on the basis of the graph, guess where the function is not differentiable. (Assume the largest possible domain.) $$ y=-|x+5| $$

Assume that the measurement of \(x\) is accurate within \(2 \% .\) In each case, determine the error \(\Delta f\) in the calculation of \(f\) and find the percentage error \(100 \frac{\Delta f}{f} .\) The quantities \(f(x)\) and the true value of \(x\) are given. $$ f(x)=\frac{1}{1+x}, x=4 $$

See all solutions

Recommended explanations on Biology 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.