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Chapter 3: Problem 13

Use implicit differentiation to find \(d y / d x\). \(x(y-1)^{2}=6\)

Short Answer

Expert verified
\( \frac{dy}{dx} = -\frac{y-1}{2x} \)

Step by step solution

01

Differentiate Both Sides with Respect to x

Start by differentiating both sides of the equation with respect to \(x\). For the left side, use the product rule: Let \(u = x\) and \(v = (y-1)^2\).Then, the derivative \( \frac{d}{dx}[x(y-1)^2] = u'v + uv' \).This becomes \(1 \cdot (y-1)^2 + x \cdot 2(y-1)\frac{dy}{dx}\cdot 1\) due to the chain rule.For the right side, since it is a constant, the derivative is 0.
02

Simplify and Rearrange the Equation

From Step 1, the differentiated equation is \((y-1)^2 + 2x(y-1)\frac{dy}{dx} = 0\).Rearrange terms to isolate \(\frac{dy}{dx}\):\(2x(y-1)\frac{dy}{dx} = -(y-1)^2\).
03

Solve for dy/dx

Divide both sides of the equation by \(2x(y-1)\) to solve for \(\frac{dy}{dx}\):\(\frac{dy}{dx} = -\frac{(y-1)^2}{2x(y-1)}\).Simplify the expression:\(\frac{dy}{dx} = -\frac{y-1}{2x}\).

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Key Concepts

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

Understanding the Product Rule
In calculus, the product rule is critical for differentiating expressions where two functions are multiplied together. When you have a function expressed as a product of two separate functions, like in the original exercise where we have
  • \( u = x \)
  • \( v = (y-1)^2 \)
, the product rule helps us find the derivative. The product rule states:\[ \frac{d}{dx}[u \, v] = u' \, v + u \, v' \]Here, \( u' \) is the derivative of \( u \) with respect to \( x \), and \( v' \) is the derivative of \( v \).
This rules simplifies calculating complex derivatives by breaking them down into manageable parts, allowing each function to be differentiated individually before combining them in the specified manner.
How the Chain Rule Applies
The chain rule is a fundamental tool for handling composite functions. Composite functions are functions within other functions, like our function \( (y-1)^2 \) from the original problem, making it an essential piece of implicit differentiation.To differentiate a composite function, the chain rule suggests that we
  • differentiate the outer function,
  • and multiply it by the derivative of the inner function.
Here, the outer function is \( u^2 \) and the inner function is \( u = y-1 \). So, when we apply the chain rule:\[ \frac{d}{dx}[(y-1)^2] = 2(y-1) \cdot \frac{d}{dx}[y-1] \]Implying the need for a full understanding of how derivatives of inner layers contribute to the overall derivative result.
Derivative of Constants
When differentiating expressions, constants play a straightforward role. If you remember from early calculus, the derivative of any constant, like \( 6 \) in the original exercise, is simply zero. This is because constants don't change, and so their rate of change, or derivative, is always zero.
In the example:For the equation \( x(y-1)^2 = 6 \), when differentiating the right-hand side, we find:\[ \frac{d}{dx}[6] = 0 \]This concept simplifies parts of equations during differentiation. It provides an excellent example of how constants behave under differentiation rules, ensuring that they do not affect the variables or expressions being derived.

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Most popular questions from this chapter

Use implicit differentiation to find \(d p / d x\). \(p^{3}+p+6 x=50\)

A toy manufacturer estimates the demand for a doll to be 10,000 per year. Each doll costs $$\$ 5$$ to manufacture, plus setup costs of $$\$ 800$$ for each production run. If it costs $$\$ 4$$ to store a doll for a year, how many should be manufactured at a time and how many production runs should there be to minimize costs?

Write in calculus notation: The population of a city is shrinking at the rate of 1500 per year.

Use implicit differentiation to find \(d p / d x\). \((p-1)(x+5)=24\)

Write in calculus notation: The value of my car is falling at the rate of \(\$ 2000\) per year.
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