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Chapter 3: Problem 63

(a) Find \(y^{\prime}\) by implicit differentiation. (b) Solve the equation explicitly for \(y\) and differentiate to get \(y^{\prime}\) in terms of \(x .\) (c) Check that your solutions to parts (a) and (b) are consistent by substituting the expression for \(y\) into your solution for part (a). \(x y+2 x+3 x^{2}=4\)

Short Answer

Expert verified
\( \frac{dy}{dx} = -3 - \frac{4}{x^2} \) in both methods, confirming consistency.

Step by step solution

01

Implicit Differentiation

Start by differentiating both sides of the given equation \(xy + 2x + 3x^2 = 4\) with respect to \(x\). Ensure you apply the product rule for \(xy\): \(d(xy)/dx = x \frac{dy}{dx} + y\).Differentiating, we get: \[ \frac{d}{dx}(xy) + \frac{d}{dx}(2x) + \frac{d}{dx}(3x^2) = \frac{d}{dx}(4) \] Which results in: \[ x \frac{dy}{dx} + y + 2 + 6x = 0 \].Solve for \( \frac{dy}{dx}\): \[ x \frac{dy}{dx} = -y - 6x - 2 \] \[ \frac{dy}{dx} = \frac{-y - 6x - 2}{x} \].
02

Solve for y Explicitly

Rearrange and solve the equation \(xy + 2x + 3x^2 = 4\) to express \(y\) explicitly in terms of \(x\). Start by isolating terms involving \(y\):\[ xy = 4 - 2x - 3x^2 \].Now solve for \(y\): \[ y = \frac{4 - 2x - 3x^2}{x} \].
03

Differentiate Explicit Expression for y

Differentiate the explicit expression of \(y\) to find \(\frac{dy}{dx}\). Use the quotient rule: \( \frac{d}{dx}(\frac{u}{v}) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \), where \(u = 4 - 2x - 3x^2\) and \(v = x\).So, \(\frac{du}{dx} = -2 - 6x\) and \(\frac{dv}{dx} = 1\).Applying the quotient rule:\[ \frac{dy}{dx} = \frac{x(-2 - 6x) - (4 - 2x - 3x^2)(1)}{x^2} \]Simplify the expression:\[ \frac{dy}{dx} = \frac{-2x - 6x^2 - 4 + 2x + 3x^2}{x^2} = \frac{-6x^2 - 4 + 3x^2}{x^2} \]\[ \frac{dy}{dx} = \frac{-3x^2 - 4}{x^2} = \frac{-3x^2}{x^2} - \frac{4}{x^2} \]\[ \frac{dy}{dx} = -3 - \frac{4}{x^2} \].
04

Verify Consistency

To check consistency, substitute \(y = \frac{4 - 2x - 3x^2}{x}\) into the expression for \(\frac{dy}{dx}\) found in Step 1: \(\frac{dy}{dx} = \frac{-y - 6x - 2}{x}\).Sub substitute \(y\) we have:\[ \frac{dy}{dx} = \frac{-\left(\frac{4 - 2x - 3x^2}{x}\right) - 6x - 2}{x} \]Simplify:\[ \frac{dy}{dx} = \frac{-4 + 2x + 3x^2}{x} - 6x - 2\] \[ = \frac{-4 + 2x + 3x^2 - 6x^2 - 2x}{x} \] \[ = \frac{-3x^2 - 4}{x} \] \[ = -3 - \frac{4}{x^2} \].This matches the result from Step 3, verifying consistency.

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Key Concepts

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

Product Rule
When dealing with the function \(xy\) during differentiation, it is essential to apply the product rule. The product rule is used when differentiating a product of two functions. If \(u\) and \(v\) are functions of \(x\), then the derivative of their product \(uv\) is given by \(\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}\). In our implicit differentiation,
  • \(u = x\) and \(v = y\), hence \(\frac{d(xy)}{dx} = x \frac{dy}{dx} + y\).
This technique allows us to properly account for the change in both variables simultaneously, thus obtaining the accurate derivative of the function. Without the product rule, the derivative of products would be underestimated, leading to incorrect results.
Quotient Rule
To differentiate a quotient of two functions, the quotient rule is a powerful tool. The quotient rule states that for functions \(u\) and \(v\), where \(v eq 0\),\(\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}\). This becomes particularly useful when the expression for a function is given as a fraction, as in the explicit solution for \(y\).
  • Here, \(u = 4 - 2x - 3x^2\) and \(v = x\).
  • Applying the rule ensures that each component in the numerator addresses its respective derivative, while the denominator remains \(v^2\), ensuring accuracy and completeness in the derivative calculation.
Solving Equations Explicitly
Sometimes, it is helpful to express one variable in terms of another explicitly. In our equation \(xy + 2x + 3x^2 = 4\), we isolated \(y\) to express it explicitly as a function of \(x\). This step involves basic algebraic manipulation to rearrange terms.
  • By isolating \(xy\), we obtained \(xy = 4 - 2x - 3x^2\).
  • Dividing each side by \(x\), we solved for \(y\) as \(y = \frac{4 - 2x - 3x^2}{x}\).
Dealing with explicit forms can make some mathematical processes, such as taking derivatives, more straightforward, as there is a clear functional relationship defined.
Verifying Solutions
Verification is a critical step to ensure that solutions are correct and consistent. This involves confirming that both implicit and explicit differentiation methods result in the same expression for the derivative.To verify:
  • Substitute the explicit expression for \(y\) (\(y = \frac{4 - 2x - 3x^2}{x}\)) back into the implicit form, thus checking consistency.
  • The simplification process then confirms that both methods indeed yield \(\frac{dy}{dx} = -3 - \frac{4}{x^2}\).
Consistent results affirm that calculations and methods applied were correct, giving us confidence in our solution.
Implicit Equations
Implicit equations involve relationships between variables not initially solved for one variable in terms of the other. They are ubiquitous in calculus since they allow for expressions that might be more complex or represent situations beyond simple functions.When differentiating implicitly:
  • Both sides of the equation are differentiated equally with respect to one variable (here, \(x\)).
  • Each term is treated in light of other variables, often requiring the product or chain rules.
Implicit differentiation is crucial for problems where it's difficult to isolate a variable, providing a way to deal with such complex equations easily.

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Most popular questions from this chapter

55\. Use the definition of derivative to prove that $$\lim _{x \rightarrow 0} \frac{\ln (1+x)}{x}=1$$

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