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Chapter 13: Problem 26

Find \(d y / d x\) by implicit differentiation. $$ x^{2 / 3}+y^{2 / 3}=2 $$

Short Answer

Expert verified
\(\frac{dy}{dx} = -\left(\frac{x}{y}\right)^{-1/3}\).

Step by step solution

01

Understand the problem and define implicit differentiation

The given equation \(x^{2/3} + y^{2/3} = 2\) involves both \(x\) and \(y\). We are tasked with finding \(\frac{dy}{dx}\), the derivative of \(y\) with respect to \(x\), using implicit differentiation. This technique is required because \(y\) is not isolated.
02

Differentiate with respect to x

Differentiate each term of the equation \(x^{2/3} + y^{2/3} = 2\) with respect to \(x\). Using the power rule, the derivative of the first term is \(\frac{2}{3}x^{-1/3}\). For the second term, apply the chain rule to find \(\frac{2}{3}y^{-1/3} \frac{dy}{dx}\). Differentiating the right side, the constant 2 yields 0.
03

Set up the implicit differentiation equation

Combine the derivatives from Step 2 to create the equation: \(\frac{2}{3}x^{-1/3} + \frac{2}{3}y^{-1/3} \frac{dy}{dx} = 0\).
04

Solve for dy/dx

Isolate \(\frac{dy}{dx}\) by first moving the \(\frac{2}{3}x^{-1/3}\) term to the right side, resulting in \(\frac{2}{3}y^{-1/3} \frac{dy}{dx} = -\frac{2}{3}x^{-1/3}\). Divide both sides by \(\frac{2}{3}y^{-1/3}\) to get \(\frac{dy}{dx} = -\frac{x^{-1/3}}{y^{-1/3}}\).
05

Simplify the expression for dy/dx

Simplify the right side to obtain \(\frac{dy}{dx} = -\left(\frac{x}{y}\right)^{-1/3}\). This is the implicit derivative of \(y\) with respect to \(x\).

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Key Concepts

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

Derivatives
In calculus, derivatives are a fundamental tool for understanding how functions change. When we say \( \ \frac{dy}{dx} \ \), we mean the rate of change of the variable \ \ y \ \ with respect to the variable \ \ x \ \. Essentially, it tells us how \ \ y \ \ behaves when \ \ x \ \ changes infinitesimally.
  • In explicit functions like \ \ y = f(x) \ \, finding derivatives is straightforward because \ \ y \ \ is isolated.
  • Implicit differentiation is necessary when \ \ y \ \ is intermixed with \ \ x \ \ within an equation, such as \ \ x^{2/3} + y^{2/3} = 2 \ \.
This process allows us to differentiate each term involving both \ \ x \ \ and \ \ y \ \ with respect to \ \ x \ \, resulting in a relationship that includes \ \ \frac{dy}{dx} \ \.
Chain Rule
The chain rule is an essential principle in calculus when differentiating compositions of functions. It states that if a variable \( y \) is dependent on another variable \( u \), which in turn depends on \( x \), then the derivative of \( y \) with respect to \( x \) is the derivative of \( y \) with respect to \( u \) times the derivative of \( u \) with respect to \( x \):\[ \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \].
  • In our implicit differentiation example, the term \( y^{2/3} \) needs the chain rule because \( y \) itself is a function of \( x \).
  • Therefore, its derivative with respect to \( x \) becomes \( \frac{2}{3}y^{-1/3} \cdot \frac{dy}{dx} \).
Power Rule
The power rule provides a simple way to differentiate functions of the form \( x^n \). It states that if \( f(x) = x^n \), then the derivative \( f'(x) = nx^{n-1} \). This rule makes taking derivatives straightforward when dealing with polynomial expressions. In our problem, both \( x^{2/3} \) and \( y^{2/3} \) required differentiation.
  • The derivative of \( x^{2/3} \) using the power rule is \ \frac{2}{3}x^{-1/3} \ \.
  • For \( y^{2/3} \), the power rule applies within the context of the chain rule.
Here, the power rule makes it easier to calculate each component of the function, allowing us to efficiently find the complete derivative.
Calculus
Calculus is the branch of mathematics that studies how things change. It is a powerful framework for solving a wide range of problems involving rates of change and the accumulation of quantities. The exercise demonstrates crucial calculus tools: derivatives, the chain rule, and the power rule.
  • **Derivatives** measure the rate at which one quantity changes with respect to another.
  • **The chain rule** helps articulate the relationship of composite functions when differentiating.
  • **The power rule** simplifies the process of finding derivatives for polynomial expressions.
Understanding these concepts gives a strong foundation in calculus, which is pivotal in both mathematical theory and real-world applications. Calculus lays the groundwork for advanced studies in fields like physics, engineering, and economics, where change is a central theme.

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Most popular questions from this chapter

Let \(z=f(x, y), x=r \cos \theta\), and \(y=r \sin \theta\). a. Show that \(\frac{\partial z}{\partial x}=\frac{\partial z}{\partial r} \cos \theta-\frac{\partial z}{\partial \theta} \frac{\sin \theta}{r}\) and \(\frac{\partial z}{\partial y}=\frac{\partial z}{\partial r} \sin \theta+\frac{\partial z}{\partial \theta} \frac{\cos \theta}{r}\).

When two resistors having resistances \(R_{1}\) and \(R_{2}\) are connected in parallel, the combined resistance \(R\) is given by \(R=R_{1} R_{2} /\left(R_{1}+R_{2}\right) .\) Show that $$ \frac{\partial^{2} R}{\partial R_{1}^{2}} \frac{\partial^{2} R}{\partial R_{2}^{2}}=\frac{4 R^{2}}{\left(R_{1}+R_{2}\right)^{4}} $$

Let \(g\) be a differentiable function of one variable and let \(f(x, y)=x g(y / x)\). Show that every plane tangent to the graph of \(f\) passes through the origin.

Find the minimum volume of a tetrahedron in the first octant bounded by the planes \(x=0, y=0, z=0\), and \(a\) plane tangent to the sphere \(x^{2}+y^{2}+z^{2}=1\). (Hint: If the plane is tangent to the sphere at the point \(\left(x_{0}, y_{0}, z_{0}\right)\), then the volume of the tetrahedron is \(1 /\left(6 x_{0} y_{0} z_{0}\right)\).)

Find all critical points. Determine whether each critical point yields a relative maximum value, a relative minimum value, or a saddle point. $$ k(u, v)=(u+v)^{2} $$
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