Find D2y/dx2 In Terms Of X And Y.

Finding the second derivative, denoted as d²y/dx², is a fundamental concept in calculus with wide applications in physics, engineering, and economics. It essentially tells us about the rate of change of the rate of change of a function, revealing crucial information about its concavity and inflection points. When dealing with functions defined implicitly, where y is not explicitly expressed in terms of x, the process requires a bit more care and involves implicit differentiation. This article will guide you through the process of finding d²y/dx² in terms of x and y using implicit differentiation, providing a comprehensive understanding with examples and explanations.

Understanding Implicit Differentiation

Before diving into finding the second derivative, it's crucial to grasp the concept of implicit differentiation. Unlike explicit functions (e.g., y = x² + 3x - 1), implicit functions define a relationship between x and y without isolating y on one side (e.g., x² + y² = 25).

To differentiate an implicit function with respect to x, we apply the chain rule to each term involving y. Remember that y is implicitly a function of x, so when differentiating y with respect to x, we treat it as y(x) and use the chain rule: d/dx [ y(x) ] = dy/dx.

Example:

Consider the equation x² + y² = 25. To find dy/dx, we differentiate both sides with respect to x:

d/dx (x²) + d/dx (y²) = d/dx (25)

2x + 2y (dy/dx) = 0

Solving for dy/dx, we get:

dy/dx = -x/ y

This is the first derivative, and we'll use this as a foundation for finding the second derivative.

Steps to Find d²y/dx² in Terms of x and y

Now, let's outline the general steps to find the second derivative, d²y/dx², when y is an implicit function of x:

1. Find the First Derivative (dy/dx): Differentiate the given implicit equation with respect to x, remembering to apply the chain rule to terms involving y. Solve for dy/dx.
2. Differentiate Again: Differentiate the expression for dy/dx (obtained in step 1) with respect to x. This will involve the product rule, quotient rule, and chain rule, as dy/dx itself is usually a function of both x and y. Remember that when differentiating dy/dx with respect to x, you get d²y/dx².
3. Substitute for dy/dx: The resulting expression for d²y/dx² will likely contain dy/dx. Substitute the expression you found for dy/dx in step 1 into the second derivative expression. This will eliminate dy/dx and give you d²y/dx² in terms of x and y only.
4. Simplify: Simplify the expression for d²y/dx² as much as possible. This might involve algebraic manipulation, combining terms, and factoring.

Detailed Explanation with Examples

Let's illustrate these steps with several examples:

Example 1: x² + y² = 25

This is the same equation we used earlier to demonstrate implicit differentiation. We already found that:

dy/dx = -x/ y

Now, we need to differentiate this expression with respect to x to find d²y/dx². We'll use the quotient rule:

d²y/dx² = d/dx (-x/ y) = [(y) d/dx(-x) - (-x) d/dx(y)] / (y²)

d²y/dx² = [(y)(-1) - (-x)(dy/dx)] / (y²)

d²y/dx² = [-y + x(dy/dx)] / (y²)

Now, substitute dy/dx = -x/ y into the equation:

d²y/dx² = [-y + x(-x/ y)] / (y²)

d²y/dx² = [-y - x²/ y] / (y²)

To simplify, multiply the numerator and denominator by y:

d²y/dx² = [-y² - x²] / (y³)

Notice that x² + y² = 25. Substitute this into the equation:

d²y/dx² = -25 / y³

Therefore, the second derivative d²y/dx² = -25 / y³.

Example 2: x³ + y³ = 6xy (Folium of Descartes)

This is a classic example that demonstrates the importance of the product rule in implicit differentiation.

Step 1: Find dy/dx

Differentiate both sides with respect to x:

d/dx (x³) + d/dx (y³) = d/dx (6xy)

3x² + 3y² (dy/dx) = 6 [ x (dy/dx) + y (1) ] (Using the product rule on the right side)

3x² + 3y² (dy/dx) = 6x (dy/dx) + 6y

Now, solve for dy/dx:

3y² (dy/dx) - 6x (dy/dx) = 6y - 3x²

(dy/dx) (3y² - 6x) = 6y - 3x²

dy/dx = (6y - 3x²) / (3y² - 6x)

Simplify by dividing both numerator and denominator by 3:

dy/dx = (2y - x²) / (y² - 2x)

Step 2: Find d²y/dx²

Differentiate dy/dx with respect to x. This requires the quotient rule:

d²y/dx² = d/dx [ (2y - x²) / (y² - 2x) ]

d²y/dx² = [ (y² - 2x) d/dx (2y - x²) - (2y - x²) d/dx (y² - 2x) ] / (y² - 2x)²

Now, differentiate each part carefully:

d/dx (2y - x²) = 2(dy/dx) - 2x

d/dx (y² - 2x) = 2y(dy/dx) - 2

Substitute these back into the expression for d²y/dx²:

d²y/dx² = [ (y² - 2x) (2(dy/dx) - 2x) - (2y - x²) (2y(dy/dx) - 2) ] / (y² - 2x)²

Step 3: Substitute for dy/dx

Substitute dy/dx = (2y - x²) / (y² - 2x) into the equation. This is where the algebra gets a bit intense:

d²y/dx² = [ (y² - 2x) (2((2y - x²) / (y² - 2x)) - 2x) - (2y - x²) (2y((2y - x²) / (y² - 2x)) - 2) ] / (y² - 2x)²

d²y/dx² = [ 2(2y - x²) - 2x(y² - 2x) - (2y - x²) ( (4y² - 2x²y) / (y² - 2x) - 2) ] / (y² - 2x)²

To simplify, multiply the terms inside the last parenthesis by (y² - 2x):

d²y/dx² = [ 4y - 2x² - 2xy² + 4x² - (2y - x²) ( (4y² - 2x²y - 2y² + 4x) / (y² - 2x)) ] / (y² - 2x)²

d²y/dx² = [ 4y + 2x² - 2xy² - (2y - x²) ( (2y² - 2x²y + 4x) / (y² - 2x)) ] / (y² - 2x)²

Multiply the last term in the numerator:

d²y/dx² = [ 4y + 2x² - 2xy² - (4y³ - 4x²y² + 8xy - 2x²y² + 2x⁴y - 4x³ ) / (y² - 2x) ] / (y² - 2x)²

Multiply the entire numerator and denominator by (y² - 2x) to clear the fraction:

d²y/dx² = [ (4y + 2x² - 2xy²) (y² - 2x) - (4y³ - 6x²y² + 8xy + 2x⁴y - 4x³ ) ] / (y² - 2x)³

Expand the first term:

d²y/dx² = [ (4y³ - 8xy + 2x²y² - 4x³ - 2xy⁴ + 4x²y²) - (4y³ - 6x²y² + 8xy + 2x⁴y - 4x³ ) ] / (y² - 2x*)³

Step 4: Simplify

Combine like terms:

d²y/dx² = [ 4y³ - 8xy + 2x²y² - 4x³ - 2xy⁴ + 4x²y² - 4y³ + 6x²y² - 8xy - 2x⁴y + 4x³ ] / (y² - 2x*)³

d²y/dx² = [ 12x²y² - 16xy - 2xy⁴ - 2x⁴y ] / (y² - 2x)³

Factor out -2xy:

d²y/dx² = [ -2xy (8 - 6xy + y³ + x³ ) ] / (y² - 2x)³

Remember the original equation: x³ + y³ = 6xy. Substitute this into the expression:

d²y/dx² = [ -2xy (8 - 6xy + 6xy) ] / (y² - 2x)³

d²y/dx² = [ -16xy ] / (y² - 2x)³

Therefore, the second derivative d²y/dx² = -16xy / (y² - 2x)³. This simplified form highlights the power of simplification after the substitution step.

Example 3: y sin(x) + x cos(y) = 1

This example demonstrates implicit differentiation with trigonometric functions.

Step 1: Find dy/dx

Differentiate both sides with respect to x:

d/dx (y sin(x)) + d/dx (x cos(y)) = d/dx (1)

Using the product rule for both terms:

[y' sin(x) + y cos(x)] + [cos(y) - x sin(y) y'] = 0

Here, y' represents dy/dx. Now, solve for y':

y' sin(x) - x sin(y) y' = -y cos(x) - cos(y)

y' [sin(x) - x sin(y)] = -y cos(x) - cos(y)

y' = (-y cos(x) - cos(y)) / (sin(x) - x sin(y))

dy/dx = -(y cos(x) + cos(y)) / (sin(x) - x sin(y))

Step 2: Find d²y/dx²

Differentiate dy/dx with respect to x using the quotient rule:

d²y/dx² = d/dx [ -(y cos(x) + cos(y)) / (sin(x) - x sin(y)) ]

Let u = -(y cos(x) + cos(y)) and v = (sin(x) - x sin(y))

Then, d²y/dx² = (v du/dx - u dv/dx) / v²

First, find du/dx:

du/dx = -[y' cos(x) - y sin(x) - sin(y) y']

du/dx = -[dy/dx cos(x) - y sin(x) - sin(y) dy/dx]

Next, find dv/dx:

dv/dx = cos(x) - [sin(y) + x cos(y) y']

dv/dx = cos(x) - sin(y) - x cos(y) dy/dx

Now, substitute u, v, du/dx, and dv/dx into the quotient rule formula:

d²y/dx² = { (sin(x) - x sin(y))[-dy/dx cos(x) + y sin(x) + sin(y) dy/dx] + (y cos(x) + cos(y))[cos(x) - sin(y) - x cos(y) dy/dx] } / (sin(x) - x sin(y))²

Step 3: Substitute for dy/dx

This step involves substituting the expression for dy/dx we found in step 1:

dy/dx = -(y cos(x) + cos(y)) / (sin(x) - x sin(y))

Substitute this into the expression for d²y/dx². This will result in a very complex expression. To simplify, we'll use the substitution dy/dx = y' to keep the notation manageable:

d²y/dx² = { (sin(x) - x sin(y))[-y' cos(x) + y sin(x) + sin(y) y'] + (y cos(x) + cos(y))[cos(x) - sin(y) - x cos(y) y'] } / (sin(x) - x sin(y))²

Now, substitute y' = -(y cos(x) + cos(y)) / (sin(x) - x sin(y)) :

d²y/dx² = { (sin(x) - x sin(y))[ cos(x)(y cos(x) + cos(y)) / (sin(x) - x sin(y)) + y sin(x) - sin(y)(y cos(x) + cos(y)) / (sin(x) - x sin(y)) ] + (y cos(x) + cos(y))[cos(x) - sin(y) + x cos(y) (y cos(x) + cos(y)) / (sin(x) - x sin(y))] } / (sin(x) - x sin(y))²

Step 4: Simplify

This is where the simplification becomes extremely challenging. We need to multiply through and combine like terms, and then factor out common factors. To make the expression more manageable, multiply the numerator and denominator by (sin(x) - x sin(y)) to eliminate the inner fractions:

d²y/dx² = { (sin(x) - x sin(y)) [ cos(x)(y cos(x) + cos(y)) + y sin(x)(sin(x) - x sin(y)) - sin(y)(y cos(x) + cos(y)) ] + (y cos(x) + cos(y)) [cos(x)(sin(x) - x sin(y)) - sin(y)(sin(x) - x sin(y)) + x cos(y)(y cos(x) + cos(y)) ] } / (sin(x) - x sin(y))³

Expanding and simplifying this expression further would be a lengthy process. While it's possible to simplify this further, the core concept is demonstrated: Find dy/dx, differentiate again, and then substitute dy/dx back into the equation. The algebraic complexity is often the biggest hurdle.

Key Considerations and Common Mistakes

• Chain Rule: The most common mistake is forgetting to apply the chain rule when differentiating terms involving y. Always remember that d/dx [ f(y) ] = f'(y) (dy/dx).
• Product and Quotient Rules: When differentiating dy/dx, you'll often need to use the product or quotient rule. Be meticulous in applying these rules.
• Algebraic Simplification: The algebra involved in simplifying the expression for d²y/dx² can be quite challenging. Take your time, be organized, and double-check your work. Look for opportunities to factor and combine terms.
• Substitution: Don't forget to substitute your expression for dy/dx back into the expression for d²y/dx². This is crucial for expressing d²y/dx² solely in terms of x and y.
• Notational Clarity: Use clear notation to avoid confusion, especially when dealing with multiple derivatives. Using y' and y'' can sometimes simplify the writing process, but be sure to clearly define what they represent.

Applications of the Second Derivative

Understanding and calculating the second derivative has several important applications:

• Concavity: The sign of the second derivative tells us about the concavity of the curve. If d²y/dx² > 0, the curve is concave up (like a smile). If d²y/dx² < 0, the curve is concave down (like a frown).
• Inflection Points: Inflection points are points where the concavity of the curve changes. These occur where d²y/dx² = 0 or is undefined, provided the concavity changes sign at that point.
• Optimization: In optimization problems, the second derivative test can be used to determine whether a critical point (where dy/dx = 0) is a local maximum or a local minimum. If d²y/dx² > 0 at the critical point, it's a local minimum. If d²y/dx² < 0, it's a local maximum.
• Physics: In physics, the second derivative of position with respect to time represents acceleration. Understanding acceleration is crucial in analyzing motion.

Conclusion

Finding d²y/dx² in terms of x and y using implicit differentiation is a powerful technique in calculus. It requires a solid understanding of implicit differentiation, the chain rule, product and quotient rules, and algebraic simplification. While the process can be algebraically intensive, particularly in complex examples, the underlying steps remain consistent: find the first derivative, differentiate again, substitute, and simplify. Mastering this technique opens doors to understanding the concavity of curves, finding inflection points, and solving a wide range of problems in various fields. By carefully applying the steps and practicing with examples, you can gain confidence and proficiency in finding second derivatives of implicitly defined functions. Remember to pay close attention to detail and practice regularly to master this valuable calculus skill.