/*! 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 34 $$ \text { If } y=\sin \left(x... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 34

$$ \text { If } y=\sin \left(x^{2}\right)+2 x^{3} \text {, find } d x / d y \text {. } $$

Short Answer

Expert verified
\( \frac{dx}{dy} = \frac{1}{\cos(x^2) \cdot 2x + 6x^2} \).

Step by step solution

01

Identify the Expression

We are given the expression \( y = \sin(x^2) + 2x^3 \). Our goal is to find \( \frac{dx}{dy} \), the derivative of \( x \) with respect to \( y \).
02

Differentiate with Respect to x

Differentiate both sides of the expression with respect to \( x \). Using the chain rule for \( \sin(x^2) \) and the power rule for \( 2x^3 \), we get:\[ \frac{dy}{dx} = \cos(x^2) \cdot 2x + 6x^2 \].
03

Solve for dx/dy

Recall that \( \frac{dy}{dx} \) is the derivative of \( y \) with respect to \( x \), so to find \( \frac{dx}{dy} \), you need to take the reciprocal:\[ \frac{dx}{dy} = \frac{1}{\frac{dy}{dx}} = \frac{1}{\cos(x^2) \cdot 2x + 6x^2} \].

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 91影视!

Key Concepts

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

Chain Rule
The chain rule is a powerful tool in calculus for differentiating composite functions. Think of it as a way to unwrap complex functions layer by layer. In our exercise, we are faced with differentiating \( \sin(x^2) \), which is a great example of a composite function.

Here's how the chain rule comes into play:
  • Identify the outer function and the inner function. For \( \sin(x^2) \), the outer function is \( \sin(u) \) and the inner function is \( u = x^2 \).
  • Differentiate the outer function with respect to the inner function: \( \cos(u) \).
  • Differentiate the inner function with respect to \( x \): \( 2x \).
Finally, you multiply them together to get the derivative of the composite function: \( \cos(x^2) \cdot 2x \). By breaking it into steps, the chain rule turns tricky differentiation problems into manageable parts.
Reciprocal Derivative
The reciprocal of a derivative is another key concept, especially in implicit differentiation problems. When you have \( \frac{dy}{dx} \), which represents how \( y \) changes with \( x \), sometimes you need to find the opposite: \( \frac{dx}{dy} \).

This is where the reciprocal derivative steps in: If you know \( \frac{dy}{dx} \), finding \( \frac{dx}{dy} \) is straightforward.
  • Simply take the reciprocal: \( \frac{dx}{dy} = \frac{1}{\frac{dy}{dx}} \).
In our problem, after finding \( \frac{dy}{dx} = \cos(x^2) \cdot 2x + 6x^2 \), we then find \( \frac{dx}{dy} \) by reciprocating: \( \frac{1}{\cos(x^2) \cdot 2x + 6x^2} \). This inversion is crucial in getting the derivative in the direction you need it.
Power Rule
The power rule is one of the simplest yet most important rules in differentiation. It allows us to rapidly turn powers into coefficients. For a function \( x^n \), the power rule tells us that the derivative is \( nx^{n-1} \).

This rule is straightforward and is applied to straightforward expressions like \( x^3 \). For our exercise, we apply the power rule to \( 2x^3 \).
  • Bring down the power as a coefficient: 3.
  • Multiply it by the existing coefficient: 2.
  • Decrease the power by one: \( x^{2} \).
So, differentiating \( 2x^3 \) using the power rule gives us \( 6x^2 \). This process quickly returns a derivative for polynomial terms, thus efficiently tackling part of our broader differentiation problem.

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

In Problems 23-28, an object is moving along a horizontal coordinate line according to the formula \(s=f(t)\), where \(s\), the directed distance from the origin, is in feet and \(t\) is in seconds. In each case, answer the following questions (see Examples 2 and 3). (a) What are \(v(t)\) and \(a(t)\), the velocity and acceleration, at time \(t\) ? (b) When is the object moving to the right? (c) When is it moving to the left? (d) When is its acceleration negative? (e) Draw a schematic diagram that shows the motion of the object. $$ s=2 t^{3}-6 t+5 $$

In Problems 9-16, find \(f^{\prime \prime}(2)\). $$ f(s)=s\left(1-s^{2}\right)^{3} $$

In Problems 23-28, an object is moving along a horizontal coordinate line according to the formula \(s=f(t)\), where \(s\), the directed distance from the origin, is in feet and \(t\) is in seconds. In each case, answer the following questions (see Examples 2 and 3). (a) What are \(v(t)\) and \(a(t)\), the velocity and acceleration, at time \(t\) ? (b) When is the object moving to the right? (c) When is it moving to the left? (d) When is its acceleration negative? (e) Draw a schematic diagram that shows the motion of the object. $$ s=t^{3}-9 t^{2}+24 t $$

In Problems \(1-8\), find \(d^{3} y / d x^{3}\). $$ y=\sin (7 x) $$

Sketch the graph of the circle \(x^{2}+4 x+y^{2}+3=0\) and then find equations of the two tangent lines that pass through the origin.
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Geometry

Read Explanation

Pure Maths

Read Explanation

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Decision Maths

Read Explanation
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.