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Chapter 2: Problem 5

Find dy/dx by implicit differentiation. } $$ x^{3}-2 y^{3}-y=x+2 $$

Short Answer

Expert verified
The derivative \(\frac{dy}{dx}\) of the given function is \(\frac{3x^2 - 1}{6y^2 + 1}\).

Step by step solution

01

Differentiate both sides of the equation with respect to x

We will apply the rules for differentiating powers, constants, and products, treating y as a function of x: $$ \frac{d}{dx}(x^3) - \frac{d}{dx}(2y^3) - \frac{d}{dx}(y) = \frac{d}{dx}(x) + \frac{d}{dx}(2) $$
02

Differentiate each term on both sides of the equation

Applying the rules for differentiation, remembering to use the chain rule for terms involving y: $$ (3x^2) - (6y^2)\frac{dy}{dx} - \frac{dy}{dx} = 1 + 0 $$
03

Solve for dy/dx

First, simplify the equation: $$ 3x^2 - 6y^2\frac{dy}{dx} - \frac{dy}{dx} = 1 $$ Combine the terms containing dy/dx: $$ 3x^2 - 1 = (6y^2 + 1)\frac{dy}{dx} $$ Now, solve for dy/dx by dividing both sides by (6y^2 + 1) $$ \frac{dy}{dx} = \frac{3x^2 - 1}{6y^2 + 1} $$ The derivative dy/dx is therefore: $$ \frac{dy}{dx} = \frac{3x^2 - 1}{6y^2 + 1} $$

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Key Concepts

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

Understanding Derivatives
Derivatives are fundamental to calculus and describe how a function changes as its input changes. In simple terms, a derivative tells us the rate of change of one quantity with respect to another. In mathematics, it's often written as \( \frac{dy}{dx} \), which signifies the derivative of \( y \) with respect to \( x \).
  • The derivative can be visualized as the slope of a tangent line to a curve at any given point.
  • In real-world terms, it's akin to measuring speed, where you see how distance changes over time.
  • Implicit differentiation is used when functions are not given in a specific form that is easy to differentiate.
By applying derivatives to the equation \( x^3 - 2y^3 - y = x + 2 \), we understand how \( y \) changes as \( x \) changes.
Exploring the Chain Rule
The Chain Rule is a powerful tool in calculus used for differentiating composite functions. A composite function is a function made up of two or more functions, layered one over the other.
  • In the context of implicit differentiation, we treat \( y \) as a function of \( x \), even if it isn't explicitly stated.
  • The Chain Rule allows us to find the derivative of such functions, by taking the derivative of the outer function and multiplying it by the derivative of the inner function.
  • In our equation \( x^3 - 2y^3 - y = x + 2 \), we use the chain rule to differentiate terms involving \( y \).
So, when differentiating \(-2y^3\) with respect to \( x \), we consider it as \(-2(y^3)\) and find \(-6y^2 \frac{dy}{dx}\). The \( \frac{dy}{dx} \) arises because we are applying the chain rule and acknowledging that \( y \) might change with \( x \).
Differentiation Rules Simplified
Differentiation rules provide a blueprint for finding derivatives of various types of functions. They are guidelines which show how to differentiate common functions.
  • The power rule states that the derivative of \( x^n \) is \( nx^{n-1} \).
  • Constant rule helps us understand that the derivative of a constant is zero.
  • Sum difference rules allow us to take the derivative of terms added or subtracted in a straightforward manner.
In our example, we applied these rules to each term separately. For \( x^3 \), using the power rule, it became \( 3x^2 \). On the other side, the derivative of \( x \) (a linear term) is \( 1 \) and for any constant, like \( 2 \), it's zero. These rules make differentiating manageable and clear.
Solving Equations with Implicit Differentiation
After differentiating the implicit function, we solve the equation to find the derivative of \( y \) with respect to \( x \). This process involves rearranging terms and solving for \( \frac{dy}{dx} \).
  • The aim is to isolate \( \frac{dy}{dx} \) on one side of the equation.
  • Combine terms containing \( \frac{dy}{dx} \) to simplify and solve the equation.
  • Finally, divide by the remaining coefficient to solve for \( \frac{dy}{dx} \).
For our equation, we first simplify \( 3x^2 - 6y^2\frac{dy}{dx} - \frac{dy}{dx} = 1 \) to collect like terms containing \( \frac{dy}{dx} \). Then, by factoring \( \frac{dy}{dx} \) and isolating it, we arrive at the solution \( \frac{dy}{dx} = \frac{3x^2 - 1}{6y^2 + 1} \). This final result indicates how \( y \) changes with \( x \) in the context of this equation.

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Most popular questions from this chapter

A function is called even if \(f(-x)=f(x)\) for all \(x\) in the domain of \(f\), it is called \(o d d\) if \(f(-x)=-f(x)\) for all \(x\) in the domain of \(f .\) Prove that the derivative of a differentiable even function is an odd function and that the derivative of ? differentiable odd function is an even function.

Let g denote the inverse of the function \(f\). (a) Show that the point \((a, b)\) lies on the graph of \(f .\) (b) Find \(g^{\prime}(b) .\) Suppose that \(g\) is the inverse of a function \(f .\) If \(f(2)=4\) and \(f^{\prime}(2)=3\), find \(g^{\prime}(4) .\)

Let \(g\) denote the inverse of the function \(f\). (a) Show that the point \((a, b)\) lies on the graph of \(f .\) (b) Find \(g^{\prime}(b)\) Suppose that \(g\) is the inverse of a differentiable function \(f\) and \(H=g \circ g\). If \(f(4)=3, g(4)=5, f^{\prime}(4)=\frac{1}{2}\), and \(f^{\prime}(5)=2\), find \(H^{\prime}(3) .\)

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