Find D 2 Y D X 2 \frac{d^2 Y}{d X^2} D X 2 D 2 Y By Implicit Differentiation.(a) X Y + Y 2 = 2 X Y + Y^2 = 2 X Y + Y 2 = 2 (b) X 3 Y − 4 = 0 X^3 Y^{-4} = 0 X 3 Y − 4 = 0 (c) X Cos ⁡ Y = Y X \cos Y = Y X Cos Y = Y (d) Y + Sin ⁡ Y = X Y + \sin Y = X Y + Sin Y = X

Mar 13, 2025 by ADMIN 264 views

**Implicit Differentiation: A Comprehensive Guide to Finding Second Derivatives**

Introduction

Implicit differentiation is a powerful tool in calculus that allows us to find the derivative of an implicitly defined function. In this article, we will explore how to use implicit differentiation to find the second derivative of a function. We will consider four different examples and apply the technique of implicit differentiation to find the second derivative in each case.

Implicit Differentiation: A Brief Overview

Implicit differentiation is a method of finding the derivative of an implicitly defined function. An implicitly defined function is a function that is defined in terms of an equation, rather than in terms of a single variable. The equation may involve multiple variables, and the function may be defined implicitly by the equation.

To find the derivative of an implicitly defined function, we use the chain rule and the product rule of differentiation. We differentiate both sides of the equation with respect to the independent variable, and then solve for the derivative of the dependent variable.

Example (a): $x y + y^2 = 2$

To find the second derivative of the function defined by the equation $x y + y^2 = 2$ , we will use implicit differentiation.

Step 1: Differentiate both sides of the equation with respect to x

We differentiate both sides of the equation with respect to x using the chain rule and the product rule of differentiation.

\frac{d}{dx} (x y + y^2) = \frac{d}{dx} (2)

Using the product rule, we get:

\frac{d}{dx} (x y) + \frac{d}{dx} (y^2) = \frac{d}{dx} (2)

y + x \frac{dy}{dx} + 2 y \frac{dy}{dx} = 0

Step 2: Solve for $\frac{dy}{dx}$

We solve for $\frac{dy}{dx}$ by isolating it on one side of the equation.

x \frac{dy}{dx} + 2 y \frac{dy}{dx} = -y

\frac{dy}{dx} (x + 2 y) = -y

\frac{dy}{dx} = \frac{-y}{x + 2 y}

Step 3: Find the second derivative

To find the second derivative, we differentiate the first derivative with respect to x.

\frac{d^2 y}{dx^2} = \frac{d}{dx} \left(\frac{-y}{x + 2 y}\right)

Using the quotient rule, we get:

\frac{d^2 y}{dx^2} = \frac{(x + 2 y) \frac{dy}{dx} (-1) - (-y) (1 + 2 \frac{dy}{dx})}{(x + 2 y)^2}

Substituting the expression for $\frac{dy}{dx}$ , we get:

\frac{d^2 y}{dx^2} = \frac{(x + 2 y) \left(\frac{-y}{x + 2 y}\right) (-1) - (-y) (1 + 2 \left(\frac{-y}{x + 2 y}\right))}{(x + 2 y)^2}

Simplifying the expression, we get:

\frac{d^2 y}{dx^2} = \frac{-y (x + 2 y) + y (x + 2 y) + 2 y^2}{(x + 2 y)^3}

\frac{d^2 y}{dx^2} = \frac{2 y^2}{(x + 2 y)^3}

Example (b): $x^3 y^{-4} = 0$

To find the second derivative of the function defined by the equation $x^3 y^{-4} = 0$ , we will use implicit differentiation.

Step 1: Differentiate both sides of the equation with respect to x

We differentiate both sides of the equation with respect to x using the chain rule and the product rule of differentiation.

\frac{d}{dx} (x^3 y^{-4}) = \frac{d}{dx} (0)

Using the product rule, we get:

\frac{d}{dx} (x^3) y^{-4} + x^3 \frac{d}{dx} (y^{-4}) = 0

3 x^2 y^{-4} - 4 x^3 y^{-5} \frac{dy}{dx} = 0

Step 2: Solve for $\frac{dy}{dx}$

We solve for $\frac{dy}{dx}$ by isolating it on one side of the equation.

- 4 x^3 y^{-5} \frac{dy}{dx} = - 3 x^2 y^{-4}

\frac{dy}{dx} = \frac{3 x^2 y^{-4}}{4 x^3 y^{-5}}

\frac{dy}{dx} = \frac{3 x^2}{4 x^3 y}

Step 3: Find the second derivative

To find the second derivative, we differentiate the first derivative with respect to x.

\frac{d^2 y}{dx^2} = \frac{d}{dx} \left(\frac{3 x^2}{4 x^3 y}\right)

Using the quotient rule, we get:

\frac{d^2 y}{dx^2} = \frac{(4 x^3 y) \frac{dy}{dx} (3 x^2) - (3 x^2) (4 x^3 y + 3 x^2 \frac{dy}{dx})}{(4 x^3 y)^2}

Substituting the expression for $\frac{dy}{dx}$ , we get:

\frac{d^2 y}{dx^2} = \frac{(4 x^3 y) \left(\frac{3 x^2}{4 x^3 y}\right) (3 x^2) - (3 x^2) (4 x^3 y + 3 x^2 \left(\frac{3 x^2}{4 x^3 y}\right))}{(4 x^3 y)^2}

Simplifying the expression, we get:

\frac{d^2 y}{dx^2} = \frac{9 x^4 - 12 x^4 y - 9 x^4}{16 x^6 y^2}

\frac{d^2 y}{dx^2} = \frac{-3 x^4}{16 x^6 y^2}

Example (c): $x \cos y = y$

To find the second derivative of the function defined by the equation $x \cos y = y$ , we will use implicit differentiation.

Step 1: Differentiate both sides of the equation with respect to x

We differentiate both sides of the equation with respect to x using the chain rule and the product rule of differentiation.

\frac{d}{dx} (x \cos y) = \frac{d}{dx} (y)

Using the product rule, we get:

\cos y + x (-\sin y) \frac{dy}{dx} = \frac{dy}{dx}

Step 2: Solve for $\frac{dy}{dx}$

We solve for $\frac{dy}{dx}$ by isolating it on one side of the equation.

x (-\sin y) \frac{dy}{dx} - \frac{dy}{dx} = -\cos y

\frac{dy}{dx} (x (-\sin y) - 1) = -\cos y

\frac{dy}{dx} = \frac{-\cos y}{x (-\sin y) - 1}

Step 3: Find the second derivative

To find the second derivative, we differentiate the first derivative with respect to x.

\frac{d^2 y}{dx^2} = \frac{d}{dx} \left(\frac{-\cos y}{x (-\sin y) - 1}\right)

Using the quotient rule, we get:

\frac{d^2 y}{dx^2} = \frac{(x (-\sin y) - 1) \frac{dy}{dx} (-\sin y) - (-\cos y) (x (-\cos y) \frac{dy}{dx} + \sin y)}{(x (-\sin y) - 1)^2}

Substituting the expression for $\frac{dy}{dx}$ , we get:

\frac{d^2 y}{dx^2} = \frac{(x (-\sin y) - 1) \left(\frac{-\cos y}{x (-\sin y) - 1}\right) (-\sin y) - (-\cos y) (x (-\cos y) \left(\frac{-\cos y}{x (-\sin y) - 1}\right) + \sin y)}{(x (-\sin y) - 1)^2}

Simplifying the expression, we get:

Q&A: Implicit Differentiation

Q: What is implicit differentiation?

A: Implicit differentiation is a method of finding the derivative of an implicitly defined function. An implicitly defined function is a function that is defined in terms of an equation, rather than in terms of a single variable.

Q: How do I use implicit differentiation to find the second derivative of a function?

A: To use implicit differentiation to find the second derivative of a function, you need to follow these steps:

Differentiate both sides of the equation with respect to the independent variable.
Solve for the derivative of the dependent variable.
Differentiate the first derivative with respect to the independent variable to find the second derivative.

Q: What are some common mistakes to avoid when using implicit differentiation?

A: Some common mistakes to avoid when using implicit differentiation include:

Not differentiating both sides of the equation with respect to the independent variable.
Not solving for the derivative of the dependent variable.
Not differentiating the first derivative with respect to the independent variable to find the second derivative.
Not checking for extraneous solutions.

Q: How do I check for extraneous solutions when using implicit differentiation?

A: To check for extraneous solutions when using implicit differentiation, you need to substitute the expression for the derivative of the dependent variable back into the original equation and check if it is true.

Q: Can I use implicit differentiation to find the derivative of a function that is defined in terms of a single variable?

A: No, you cannot use implicit differentiation to find the derivative of a function that is defined in terms of a single variable. Implicit differentiation is used to find the derivative of an implicitly defined function, which is a function that is defined in terms of an equation.

Q: What are some real-world applications of implicit differentiation?

A: Some real-world applications of implicit differentiation include:

Finding the rate of change of a quantity that is defined in terms of an equation.
Modeling the behavior of a system that is defined in terms of an equation.
Finding the maximum or minimum value of a function that is defined in terms of an equation.

Example Solutions

Example (a): $x y + y^2 = 2$

To find the second derivative of the function defined by the equation $x y + y^2 = 2$ , we will use implicit differentiation.

Step 1: Differentiate both sides of the equation with respect to x

We differentiate both sides of the equation with respect to x using the chain rule and the product rule of differentiation.

Introduction

Implicit Differentiation: A Brief Overview

Example (a): xy+y2=2x y + y^2 = 2xy+y2=2

Step 1: Differentiate both sides of the equation with respect to x

Step 2: Solve for dydx\frac{dy}{dx}dxdy​

Step 3: Find the second derivative

Example (b): x3y−4=0x^3 y^{-4} = 0x3y−4=0

Step 1: Differentiate both sides of the equation with respect to x

Step 2: Solve for dydx\frac{dy}{dx}dxdy​

Step 3: Find the second derivative

Example (c): xcos⁡y=yx \cos y = yxcosy=y

Step 1: Differentiate both sides of the equation with respect to x

Step 2: Solve for dydx\frac{dy}{dx}dxdy​

Step 3: Find the second derivative

Q&A: Implicit Differentiation

Q: What is implicit differentiation?

Q: How do I use implicit differentiation to find the second derivative of a function?

Q: What are some common mistakes to avoid when using implicit differentiation?

Q: How do I check for extraneous solutions when using implicit differentiation?

Q: Can I use implicit differentiation to find the derivative of a function that is defined in terms of a single variable?

Q: What are some real-world applications of implicit differentiation?

Example Solutions

Example (a): xy+y2=2x y + y^2 = 2xy+y2=2

Step 1: Differentiate both sides of the equation with respect to x

Example (a): $x y + y^2 = 2$

Step 2: Solve for $\frac{dy}{dx}$

Example (b): $x^3 y^{-4} = 0$

Step 2: Solve for $\frac{dy}{dx}$

Example (c): $x \cos y = y$

Step 2: Solve for $\frac{dy}{dx}$

Example (a): $x y + y^2 = 2$