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Chapter 3: Problem 13

Use implicit differentiation to find \(d y / d x\). \begin{equation} y \sin \left(\frac{1}{y}\right)=1-x y \end{equation}

Short Answer

Expert verified
\( \frac{dy}{dx} = \frac{-y}{\sin\left(\frac{1}{y}\right) - \frac{1}{y} \cos\left(\frac{1}{y}\right) - x} \)

Step by step solution

01

Differentiate Both Sides with Respect to x

The given function is \( y \sin\left(\frac{1}{y}\right) = 1 - xy \). Apply implicit differentiation by differentiating both sides with respect to \( x \). Remember to use the product rule and chain rule where necessary. Differentiating the left side involves the product rule: first for \( y \) and \( \sin\left(\frac{1}{y}\right) \). On the right side, apply the product rule on \( xy \).
02

Differentiate the Left Side

Differentiating \( y \sin\left(\frac{1}{y}\right) \):- Using the product rule, we have \( \frac{d}{dx}[y] \cdot \sin\left(\frac{1}{y}\right) + y \cdot \frac{d}{dx} [\sin\left(\frac{1}{y}\right)] \).- \( \frac{d}{dx}[y] = \frac{dy}{dx} \).- Apply the chain rule on \( \sin\left(\frac{1}{y}\right) \): \( \cos\left(\frac{1}{y}\right) \cdot \left(-\frac{1}{y^2}\right) \cdot \frac{dy}{dx} \).- Result: \( \frac{dy}{dx} \sin\left(\frac{1}{y}\right) + y \cos\left(\frac{1}{y}\right) \left(-\frac{1}{y^2}\right) \frac{dy}{dx} \).
03

Differentiate the Right Side

Differentiating \( 1 - xy \):- The derivative of 1 is 0.- For \( xy \), apply the product rule: \( x \cdot \frac{d}{dx}[y] + y \cdot \frac{d}{dx}[x] \).- \( \frac{d}{dx}[y] = \frac{dy}{dx} \) and \( \frac{d}{dx}[x] = 1 \).- Result: \( -x \frac{dy}{dx} - y \).
04

Combine and Solve for \( \frac{dy}{dx} \)

Combine all differentiated parts into the equation:\[ \frac{dy}{dx} \sin\left(\frac{1}{y}\right) + y \cos\left(\frac{1}{y}\right) \left(-\frac{1}{y^2}\right) \frac{dy}{dx} = -x \frac{dy}{dx} - y \]Factor out \( \frac{dy}{dx} \):\[ \frac{dy}{dx} \left( \sin\left(\frac{1}{y}\right) + \frac{x}{y} \cos\left(\frac{1}{y}\right) \right) = -y \]Solve for \( \frac{dy}{dx} \):\[ \frac{dy}{dx} = \frac{-y}{\sin\left(\frac{1}{y}\right) + \frac{x}{y} \cos\left(\frac{1}{y}\right)} \]
05

Simplify the Expression

The expression is simplified by ensuring all terms are accounted. Simplifying gives:\[ \frac{dy}{dx} = \frac{-y}{\sin\left(\frac{1}{y}\right) - \frac{1}{y} \cos\left(\frac{1}{y}\right) - x} \]

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Key Concepts

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

Chain Rule
The chain rule is an essential concept in calculus, particularly when dealing with composite functions. It allows us to differentiate a function that is composed of other functions. In an implicit differentiation context, like the one we have here, the chain rule becomes crucial when dealing with nested functions such as \( \sin\left(\frac{1}{y}\right) \).

To apply the chain rule, consider the derivative of \( \sin\left(\frac{1}{y}\right) \):
  • Identify the outer function and inner function. Here, \( \sin \) is the outer function, and \( \frac{1}{y} \) is the inner function.
  • Differentiate the outer function: \( \cos\left(\frac{1}{y}\right) \), keeping the inner function unchanged.
  • Multiply by the derivative of the inner function: \( -\frac{1}{y^2} \cdot \frac{dy}{dx} \).
This application gives us the derivative expression needed in step 2 of the solution.
Product Rule
The product rule is necessary when differentiating products of functions. In our differentiation problem, we frequently encounter terms like \( y \sin\left(\frac{1}{y}\right) \) and \( xy \), each needing the product rule.

The product rule states that the derivative of two multiplied functions \( u(x) \cdot v(x) \) is \( u'(x) \cdot v(x) + u(x) \cdot v'(x) \).
  • For \( y \sin\left(\frac{1}{y}\right) \), \( u(x) = y \) and \( v(x) = \sin\left(\frac{1}{y}\right) \). Thus: \( \frac{dy}{dx} \cdot \sin\left(\frac{1}{y}\right) + y \cdot \) derivative of \( \sin\left(\frac{1}{y}\right) \) as derived using the chain rule.
  • For \( xy \), \( u(x) = x \) and \( v(x) = y \), the product rule gives \( x \cdot \frac{dy}{dx} + y \cdot 1 \).
This use of the product rule simplifies the differentiation process, allowing us to discriminate between terms more easily.
Derivatives
Derivatives represent the rate of change of a function. When using implicit differentiation, we find derivatives with respect to one variable even if the function involves others.

In our particular problem, derivatives are taken with respect to \( x \) even though the equation includes both \( x \) and \( y \).
  • The derivative of \( y \) with respect to \( x \) is \( \frac{dy}{dx} \). This is true for any direct appearances of \( y \) in the equation.
  • The derivative of \( x \) is simply 1, because \( \frac{d}{dx}[x] \) equals 1.
  • For constant terms like 1, the derivative is 0 as constants don鈥檛 change with respect to \( x \).
Understanding these derivatives aids in successfully manipulating and solving equations using calculus, like finding \( \frac{dy}{dx} \) in the given problem.

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Most popular questions from this chapter

$$Find the tangent to y=((x-1) /(x+1))^{2} at x=0$$

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