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Chapter 3: Problem 31

Second derivatives Find \(\frac{d^{2} y}{d x^{2}}\). $$x+y^{2}=1$$

Short Answer

Expert verified
Answer: The second derivative of y with respect to x is \(\frac{d^2y}{dx^2} = \frac{1}{4y^3}\).

Step by step solution

01

Implicit differentiation to find the first derivative (dy/dx)

To find the first derivative, we will differentiate both sides of the equation with respect to x, using implicit differentiation: $$\frac{d}{dx}(x+y^2) = \frac{d}{dx}(1)$$ Differentiating x and y^2 with respect to x, we get: $$\frac{dx}{dx} + 2y\frac{dy}{dx} = 0$$ Now, simplify: $$1 + 2y\frac{dy}{dx} = 0$$ Solve for dy/dx: $$\frac{dy}{dx} = -\frac{1}{2y}$$
02

Implicit differentiation to find the second derivative (d虏y/dx虏)

Now that we have the first derivative, we can differentiate it again to find the second derivative. Again, we will use implicit differentiation: $$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(-\frac{1}{2y}\right)$$ Applying the chain rule to the derivative: $$\frac{d^2y}{dx^2} = -\frac{1}{2}\frac{d}{dx}(y^{-1})$$ $$\frac{d^2y}{dx^2} = -\frac{1}{2}(-1)y^{-2}\frac{dy}{dx}$$ Now, substitute the expression for dy/dx we found in Step 1: $$\frac{d^2y}{dx^2} = -\frac{1}{2}(-1)y^{-2}(-\frac{1}{2y})$$ Simplify the expression: $$\frac{d^2y}{dx^2} = \frac{1}{4y^3}$$ The second derivative of y with respect to x is: $$\frac{d^2y}{dx^2} = \frac{1}{4y^3}$$

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Key Concepts

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

Implicit Differentiation
Implicit differentiation is a technique used in calculus when we have equations involving both independent and dependent variables that are intertwined and not easily separated. Let's consider the equation given in the exercise: \(x + y^2 = 1\). Here, \(y\) is a function of \(x\), although not explicitly given as \(y = f(x)\).
To find the derivative \(\frac{dy}{dx}\), we differentiate both sides with respect to \(x\) using implicit differentiation. Remember:
  • Differentiate each term individually.
  • For \(y^2\), treat \(y\) as a function of \(x\). Apply the chain rule which we'll discuss later, resulting in \(2y \frac{dy}{dx}\).
  • Equate to the derivative of the right-hand side, which is 0 since the derivative of a constant (1) is 0.
After simplification, we solve for \(\frac{dy}{dx} = -\frac{1}{2y}\). By implicitly differentiating, we treat \(y\) as an expression dependent on \(x\), allowing us to find derivatives even when \(y\) isn't isolated.
Second Derivative
The second derivative provides information about the curvature or concavity of a function. After finding the first derivative \(\frac{dy}{dx} = -\frac{1}{2y}\), we move on to find the second derivative \(\frac{d^2y}{dx^2}\).
Differentiating the first derivative, we apply implicit differentiation again. This time, it becomes slightly more complex because \(\frac{1}{2y}\) is a term in itself apart from \(y\). We write:
  • \(\frac{d}{dx}\left(-\frac{1}{2y}\right)\)
  • Recognize \(y^{-1}\) as a functional relationship and observe the negative exponent.
  • Differentiate with respect to \(x\), using the power and chain rules to manage \(y^{-1}\). This results in \(-\frac{1}{2}(-1)y^{-2}\cdot \frac{dy}{dx}\).
By substituting \(\frac{dy}{dx}\) from earlier into this expression, we arrive at the second derivative \(\frac{d^2y}{dx^2} = \frac{1}{4y^3}\). This derivative helps us understand how \(y\) bends as \(x\) alters.
Chain Rule
The chain rule is an essential tool in calculus for finding derivatives of composite functions. In implicit differentiation, it plays a critical role when dealing with terms where the dependent variable is raised to a power, like \(y^2\).
The chain rule states that if you have a composite function \(f(g(x))\), then the derivative \(f'(g(x)) \cdot g'(x)\) is used.
  • In terms of our implicit differentiation, \(y\) is a function of \(x\), even if it's not explicitly shown.
  • When differentiating \(y^2\), write \(2y\cdot \frac{dy}{dx}\) because of the chain rule.
  • This process ensures that all instances of \(y\), when expressed as functions reliant on \(x\), are treated appropriately.
Using the chain rule with implicit differentiation gives us the ability to "link" all relevant rate changes together. It's critical for finding both first and second derivatives when variables aren't isolated.

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