Mastering Implicit Differentiation A Comprehensive Guide With Examples And Practice Problems

In the realm of calculus, implicit differentiation stands as a powerful technique for finding the derivative of a function that is not explicitly defined. Unlike explicit functions where one variable is directly expressed in terms of another (e.g., y = f(x)), implicit functions involve equations where the relationship between variables is intertwined (e.g., f(x, y) = 0). This comprehensive guide delves into the intricacies of implicit differentiation, providing a step-by-step approach to solving various problems and enhancing your understanding of this essential calculus concept.

Understanding Implicit Functions and Differentiation

To truly master implicit differentiation, a solid grasp of what implicit functions are and why we need specialized differentiation techniques is crucial. An implicit function is essentially an equation that relates two or more variables but doesn't explicitly solve for one variable in terms of the others. Think of it as a hidden relationship waiting to be uncovered. For example, the equation x² + y² = 25 represents a circle, where y is implicitly defined as a function of x. We cannot simply isolate y and express it as y = f(x) for the entire circle because the square root introduces both positive and negative solutions. This is where implicit differentiation shines, allowing us to find dy/dx without explicitly solving for y. The need for implicit differentiation arises because many real-world relationships are naturally expressed in implicit forms. Equations describing curves, surfaces, and even physical laws often link variables in complex ways, making explicit solutions impractical or impossible to obtain. Furthermore, implicit differentiation can sometimes be more efficient than solving for y and then differentiating, particularly when dealing with complicated equations. The beauty of the technique lies in its ability to treat y as a function of x directly within the equation, allowing us to find the rate of change of y with respect to x in its implicit form. The core principle behind implicit differentiation is the chain rule. Remember that the chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. When differentiating a term involving y with respect to x, we're essentially dealing with a composite function where y is a function of x. Therefore, we must apply the chain rule, differentiating the term with respect to y and then multiplying by dy/dx. For instance, the derivative of y² with respect to x is 2y (dy/dx). This additional dy/dx term is the hallmark of implicit differentiation and serves as a crucial piece of the puzzle when solving for the desired derivative. Understanding this concept is vital as it forms the backbone of the entire process. When differentiating implicitly, you're essentially acknowledging the hidden dependency of y on x and using the chain rule to account for this relationship. Without this recognition, the differentiation process will be incomplete and lead to incorrect results. Therefore, always remember to apply the chain rule whenever you encounter a term involving y while differentiating with respect to x.

Step-by-Step Guide to Implicit Differentiation

Implicit differentiation might seem daunting at first, but by following a systematic approach, it becomes a manageable and powerful tool. Let's break down the process into clear, actionable steps. Firstly, the crucial first step is to differentiate both sides of the equation with respect to x. This is the foundational action that sets the stage for the entire process. Remember, an equation is a balance, and whatever operation you perform on one side, you must perform on the other to maintain that balance. When differentiating, apply the appropriate rules – power rule, product rule, quotient rule, chain rule – to each term on both sides. This might involve differentiating terms involving x, terms involving y, and potentially terms involving both. The second step involves carefully applying the chain rule whenever you differentiate a term containing y. This is the heart of implicit differentiation, as it acknowledges that y is implicitly a function of x. Every time you encounter a y term, you differentiate it with respect to y as usual, but then you must multiply by dy/dx. For example, the derivative of sin(y) with respect to x is cos(y) * (dy/dx). Similarly, the derivative of y³ with respect to x is 3y² * (dy/dx). Don't forget this crucial step, as omitting it will lead to incorrect results. Following the differentiation, the third step focuses on algebraic manipulation. The goal here is to isolate dy/dx on one side of the equation. This often involves rearranging terms, factoring out dy/dx, and performing other algebraic operations. Treat dy/dx as a variable you're trying to solve for. Combine all terms containing dy/dx on one side and all other terms on the other side. Once you have all the dy/dx terms together, factor dy/dx out as a common factor. The final step is to divide both sides of the equation by the expression that's multiplying dy/dx. This isolates dy/dx and gives you the derivative of y with respect to x. The resulting expression for dy/dx will often involve both x and y, reflecting the implicit nature of the original function. This is perfectly normal and expected in implicit differentiation. The derivative provides the slope of the tangent line to the curve defined by the implicit equation at a given point (x, y). By following these four steps diligently – differentiate both sides, apply the chain rule, isolate dy/dx, and solve for dy/dx – you'll be well-equipped to tackle a wide range of implicit differentiation problems.

Solving Implicit Differentiation Problems: Examples

To solidify your understanding of implicit differentiation, let's work through several examples. These examples will showcase the application of the step-by-step guide we discussed earlier and highlight common scenarios you might encounter. The examples in this guide cover a range of common implicit differentiation problems, from trigonometric functions to polynomials and composite functions. By studying these examples, you'll gain a practical understanding of how to apply the techniques and avoid common pitfalls. Let's tackle our first example: sin(y) + x² + 4y = cos(x). Following our step-by-step guide, the first step is to differentiate both sides of the equation with respect to x. Applying the derivative rules, we get: cos(y) (dy/dx) + 2x + 4(dy/dx) = -sin(x). Notice the application of the chain rule to the sin(y) and 4y terms, resulting in the (dy/dx) factor. Now, step two involves isolating dy/dx. We group the terms containing dy/dx on one side: cos(y) (dy/dx) + 4(dy/dx) = -sin(x) - 2x. Next, we factor out dy/dx: (dy/dx) [cos(y) + 4] = -sin(x) - 2x. Finally, we divide both sides by (cos(y) + 4) to solve for dy/dx: dy/dx = (-sin(x) - 2x) / (cos(y) + 4). This is our derivative. For our second example, let's consider 3xy² + cos(y²) = 2x³ + 5. Differentiating both sides with respect to x, we need to apply the product rule to the 3xy² term and the chain rule to both cos(y²) and 2x³ terms. This gives us: 3y² + 3x(2y)(dy/dx) - sin(y²)(2y)(dy/dx) = 6x². Simplifying and rearranging, we get: 3y² + 6xy(dy/dx) - 2y sin(y²)(dy/dx) = 6x². Now, we isolate dy/dx terms: 6xy(dy/dx) - 2y sin(y²)(dy/dx) = 6x² - 3y². Factoring out dy/dx: (dy/dx) [6xy - 2y sin(y²)] = 6x² - 3y². Finally, we divide to solve for dy/dx: dy/dx = (6x² - 3y²) / [6xy - 2y sin(y²)]. This is another example highlighting the importance of the product and chain rules in implicit differentiation. Let's examine our third example: 5x² - x³ sin(y) + 5xy = 10. This example involves both the product rule and the chain rule, making it a good test of our understanding. Differentiating both sides with respect to x, we get: 10x - [3x² sin(y) + x³ cos(y) (dy/dx)] + [5y + 5x(dy/dx)] = 0. Notice the product rule applied to both -x³ sin(y) and 5xy terms. Now, distribute the negative sign and rearrange: 10x - 3x² sin(y) - x³ cos(y) (dy/dx) + 5y + 5x(dy/dx) = 0. Isolate dy/dx terms: -x³ cos(y) (dy/dx) + 5x(dy/dx) = -10x + 3x² sin(y) - 5y. Factor out dy/dx: (dy/dx) [-x³ cos(y) + 5x] = -10x + 3x² sin(y) - 5y. Finally, solve for dy/dx: dy/dx = (-10x + 3x² sin(y) - 5y) / [-x³ cos(y) + 5x]. These examples demonstrate the core principles of implicit differentiation and the importance of applying the chain and product rules correctly. Practice with these and similar problems will build your confidence and proficiency in this technique.

Common Mistakes to Avoid in Implicit Differentiation

While implicit differentiation is a powerful technique, it's also prone to certain common errors. Being aware of these pitfalls can help you avoid them and ensure accurate results. One of the most frequent mistakes is forgetting the chain rule when differentiating terms involving y. Remember, y is implicitly a function of x, so every time you differentiate a y term with respect to x, you must multiply by dy/dx. For example, the derivative of y³ with respect to x is 3y² (dy/dx), not just 3y². Omitting this crucial (dy/dx) factor will lead to an incorrect derivative. Another common error occurs when applying the product rule. The product rule states that the derivative of uv is u'v + uv'. When dealing with implicit functions, you might have terms like xy or x²y, where both x and y are variables. Make sure to apply the product rule correctly, differentiating each part of the product and including the appropriate (dy/dx) term when differentiating y. A third common mistake involves algebraic errors when isolating dy/dx. After differentiating both sides of the equation, you'll need to rearrange terms, factor out dy/dx, and divide. This algebraic manipulation can be tricky, especially with complex equations. Double-check each step to avoid errors in arithmetic or sign changes. A final pitfall is failing to simplify the final expression for dy/dx. While it's not always necessary to simplify, a simplified expression is often easier to work with and interpret. Look for opportunities to factor, cancel terms, or combine like terms. Simplification can also help you identify potential errors in your calculations. In summary, to avoid mistakes in implicit differentiation, always remember the chain rule, apply the product rule carefully, double-check your algebra, and simplify your final answer whenever possible. By being mindful of these common errors, you can significantly improve your accuracy and confidence in solving implicit differentiation problems.

Practice Problems and Solutions

To truly master implicit differentiation, consistent practice is key. Working through a variety of problems will help you solidify your understanding of the steps involved and identify areas where you might need further clarification. This section provides a set of practice problems with detailed solutions to guide your learning. Let's start with some fundamental problems. Problem 1: Find dy/dx for the equation x² + y² = 25. This classic equation represents a circle, and it's a great starting point for understanding implicit differentiation. Differentiating both sides with respect to x, we get: 2x + 2y(dy/dx) = 0. Now, isolate the dy/dx term: 2y(dy/dx) = -2x. Finally, solve for dy/dx: dy/dx = -x/y. This result tells us the slope of the tangent line to the circle at any point (x, y) on the circle. Problem 2: Find dy/dx for the equation y sin(x) + x cos(y) = 0. This problem involves trigonometric functions and the product rule, adding a bit more complexity. Differentiating both sides with respect to x, we apply the product rule to both y sin(x) and x cos(y): [ (dy/dx) sin(x) + y cos(x) ] + [ cos(y) - x sin(y) (dy/dx) ] = 0. Now, group the terms containing dy/dx: (dy/dx) sin(x) - x sin(y) (dy/dx) = -y cos(x) - cos(y). Factor out dy/dx: (dy/dx) [ sin(x) - x sin(y) ] = -y cos(x) - cos(y). Finally, solve for dy/dx: dy/dx = [ -y cos(x) - cos(y) ] / [ sin(x) - x sin(y) ]. This derivative represents the slope of the tangent line to the curve defined by the equation at any point (x, y). Problem 3: Find dy/dx for the equation tan(xy) = x² + y². This problem introduces a composite function and requires careful application of the chain rule. Differentiating both sides with respect to x, we get: sec²(xy) [ y + x(dy/dx) ] = 2x + 2y(dy/dx). Distribute the sec²(xy) term: y sec²(xy) + x sec²(xy) (dy/dx) = 2x + 2y(dy/dx). Group the dy/dx terms: x sec²(xy) (dy/dx) - 2y(dy/dx) = 2x - y sec²(xy). Factor out dy/dx: (dy/dx) [ x sec²(xy) - 2y ] = 2x - y sec²(xy). Finally, solve for dy/dx: dy/dx = [ 2x - y sec²(xy) ] / [ x sec²(xy) - 2y ]. By working through these practice problems and their solutions, you can gain confidence in your ability to apply the techniques of implicit differentiation to a variety of equations. Remember to focus on understanding the underlying concepts and the proper application of the differentiation rules.

Applications of Implicit Differentiation

Implicit differentiation isn't just a theoretical exercise; it has numerous practical applications in various fields. Understanding these applications can provide a deeper appreciation for the power and versatility of this technique. One key application of implicit differentiation is in finding the equations of tangent lines to curves defined implicitly. Recall that the derivative dy/dx gives the slope of the tangent line at a given point (x, y) on the curve. Once you've found dy/dx using implicit differentiation, you can plug in the coordinates of a specific point to find the slope at that point. Then, using the point-slope form of a line, you can easily write the equation of the tangent line. This is particularly useful for curves that are difficult or impossible to express explicitly as y = f(x). Another important application is in related rates problems. These problems involve finding the rate of change of one quantity in terms of the rate of change of another quantity, where the quantities are related by an implicit equation. For example, consider a spherical balloon being inflated. The volume V and the radius r of the balloon are related by the equation V = (4/3)πr³. If we know the rate at which the volume is increasing (dV/dt), we can use implicit differentiation to find the rate at which the radius is increasing (dr/dt). Differentiating both sides of the equation with respect to time t, we get: dV/dt = 4πr² (dr/dt). Then, we can solve for dr/dt in terms of dV/dt and r. Related rates problems arise in many areas of science and engineering, making implicit differentiation a valuable tool for solving them. Beyond tangent lines and related rates, implicit differentiation is also used in optimization problems, curve sketching, and various other calculus applications. It's a fundamental technique that underpins many advanced concepts in mathematics and its applications. By mastering implicit differentiation, you'll not only gain a powerful tool for solving specific problems but also develop a deeper understanding of the relationships between variables and their rates of change. This understanding is essential for success in calculus and related fields.

Conclusion

Implicit differentiation is a cornerstone of calculus, enabling us to tackle equations where variables are intertwined. This guide has walked you through the core concepts, a step-by-step solution process, common pitfalls to avoid, and practical applications. With consistent practice and a solid understanding of the principles outlined here, you'll be well-equipped to confidently handle implicit differentiation problems. Remember that implicit differentiation is a powerful tool with wide-ranging applications. By mastering this technique, you'll not only enhance your calculus skills but also gain a deeper understanding of the relationships between variables and their rates of change. So, keep practicing, keep exploring, and keep pushing the boundaries of your mathematical knowledge. The world of calculus awaits your discoveries!

1. Basic Implicit Differentiation

a) Find dy/dx for y

To find dy/dx for y, we differentiate both sides of the equation y = y with respect to x. The derivative of y with respect to x is simply dy/dx. Therefore, dy/dx = dy/dx. This might seem trivial, but it reinforces the fundamental concept of differentiation. In this case, dy/dx is expressed implicitly as a function of x, even though it doesn't involve x explicitly. This is a basic example that highlights the nature of implicit differentiation, where we find the derivative of a function defined implicitly rather than explicitly. This problem serves as a starting point for understanding how implicit differentiation works, laying the groundwork for more complex problems. Understanding this simple case helps in grasping the more intricate applications of implicit differentiation. The concept here is that dy/dx represents the rate of change of y with respect to x, and in this case, it is expressed in terms of itself. The triviality of the solution underscores the fact that we are finding the derivative of y with respect to x in an implicit form. This sets the stage for more complex scenarios where the relationship between x and y is not straightforward, requiring the full power of implicit differentiation. Recognizing the basic cases helps in building a solid foundation for tackling more challenging problems in calculus.

b) Find dy/dx for y²

To find dy/dx for y², we apply the chain rule, which is a fundamental principle in implicit differentiation. We differentiate y² with respect to y and then multiply by dy/dx. The derivative of y² with respect to y is 2y. Therefore, the derivative of y² with respect to x is 2y (dy/dx). This is a classic example of how the chain rule works in implicit differentiation, where we treat y as a function of x. The dy/dx term is crucial because it acknowledges the implicit relationship between x and y. Without this term, the differentiation would be incomplete and incorrect. This problem illustrates a key step in implicit differentiation: differentiating a term involving y with respect to x. The presence of the dy/dx term indicates that we are accounting for the rate of change of y with respect to x, which is the essence of implicit differentiation. Mastering this step is essential for solving more complex problems. By applying the chain rule correctly, we are able to find the derivative even when y is not explicitly defined as a function of x. This highlights the power and versatility of implicit differentiation as a technique for handling such situations. The 2y (dy/dx) result shows how the derivative of a y-term implicitly depends on both y and the rate of change of y with respect to x. This understanding is vital for solving more complex equations involving implicit functions.

c) Find dy/dx for sin(y)

To find dy/dx for sin(y), we again apply the chain rule. The derivative of sin(y) with respect to y is cos(y). Therefore, when differentiating with respect to x, we must multiply by dy/dx, giving us cos(y) (dy/dx). This problem further demonstrates the application of the chain rule in implicit differentiation, particularly with trigonometric functions. The cos(y) term represents the derivative of the outer function (sin) evaluated at the inner function (y), and the dy/dx term accounts for the derivative of the inner function with respect to x. This step is essential in ensuring the correctness of the implicit differentiation process. The appearance of the dy/dx term is a clear indicator that we are differentiating implicitly, acknowledging the dependence of y on x. This example reinforces the importance of remembering the chain rule when dealing with functions of y while differentiating with respect to x. Understanding how to handle trigonometric functions in implicit differentiation is crucial, as they often appear in more complex equations. The result, cos(y) (dy/dx), highlights how the derivative of a trigonometric function of y depends on the cosine of y and the rate of change of y with respect to x. This reinforces the concept of implicit dependence and the need for careful application of the chain rule.

d) Find dy/dx for e^(2y)

To find dy/dx for e^(2y), we once again utilize the chain rule. The derivative of e^(2y) with respect to 2y is e^(2y). The derivative of 2y with respect to y is 2. Thus, the derivative of e^(2y) with respect to y is 2e^(2y). When differentiating with respect to x, we multiply by dy/dx, giving us 2e^(2y) (dy/dx). This example showcases the chain rule applied to an exponential function within the context of implicit differentiation. We first differentiate the outer exponential function and then the inner function (2y), finally multiplying by dy/dx to account for the implicit relationship between y and x. This multi-layered application of the chain rule is a common scenario in implicit differentiation, highlighting the importance of carefully identifying the different layers of functions. The dy/dx term is crucial, as it ensures that we are capturing the rate of change of e^(2y) with respect to x, not just with respect to y. This problem reinforces the technique of handling exponential functions in implicit differentiation, which is a valuable skill for solving a wide range of problems. The result, 2e^(2y) (dy/dx), clearly shows how the derivative of the exponential function of y depends on the exponential function itself and the rate of change of y with respect to x. This further illustrates the intricacies of implicit dependence in differentiation.

e) Find dy/dx for x + y

To find dy/dx for x + y, we differentiate each term separately with respect to x. The derivative of x with respect to x is 1. The derivative of y with respect to x is dy/dx. Therefore, the derivative of x + y with respect to x is 1 + dy/dx. This example demonstrates a simple case of implicit differentiation where we differentiate a sum of terms, one involving x and the other involving y. The derivative of x is straightforward, while the derivative of y introduces the dy/dx term, highlighting the implicit nature of the differentiation. This problem emphasizes the linearity of differentiation, allowing us to differentiate each term separately and then add the results. It also reinforces the fundamental concept that the derivative of y with respect to x is dy/dx. The result, 1 + dy/dx, shows how the derivative of the sum is the sum of the derivatives, each accounting for the rate of change with respect to x. This simple example helps build confidence in applying implicit differentiation to more complex expressions. Understanding this basic case is crucial for tackling equations involving sums and differences of terms in implicit differentiation problems.

f) Find dy/dx for xy

To find dy/dx for xy, we need to apply the product rule. The product rule states that the derivative of uv is u'v + uv'. In this case, u = x and v = y. The derivative of x with respect to x is 1. The derivative of y with respect to x is dy/dx. Applying the product rule, we get: (1)(y) + (x)(dy/dx) = y + x(dy/dx). This example demonstrates the application of the product rule in implicit differentiation. When we have a product of two functions, where one or both functions implicitly depend on x, we need to use the product rule to find the derivative. The dy/dx term arises from differentiating the y term with respect to x. This problem is a classic example of how implicit differentiation differs from explicit differentiation. In explicit differentiation, we might simply have a term like x times a constant, but here, y is also a function of x, requiring the product rule. The result, y + x(dy/dx), clearly shows how the derivative of the product xy depends on both y and the rate of change of y with respect to x. This understanding is essential for solving more complex problems involving products of functions in implicit differentiation. Mastering the product rule in this context is crucial for handling equations where both x and y are intertwined.

g) Find dy/dx for y sin(x)

To find dy/dx for y sin(x), we again apply the product rule. Let u = y and v = sin(x). The derivative of y with respect to x is dy/dx. The derivative of sin(x) with respect to x is cos(x). Applying the product rule, we get: (dy/dx) sin(x) + y cos(x). This problem combines the product rule with trigonometric functions, providing a good example of a slightly more complex implicit differentiation. We need to carefully differentiate the product of y and sin(x), remembering that y is implicitly a function of x. The dy/dx term arises from differentiating the y term, as before. This example reinforces the importance of correctly applying the product rule in the context of implicit differentiation. The result, (dy/dx) sin(x) + y cos(x), highlights how the derivative of the product depends on both the derivative of y and the derivative of sin(x), as well as the original functions themselves. Understanding how to combine the product rule with trigonometric functions is a valuable skill for solving a wide range of implicit differentiation problems. This reinforces the idea that the chain rule is not the only rule at play in implicit differentiation, other rules, such as the product rule, might need to be applied as well.

h) Find dy/dx for y sin(y)

To find dy/dx for y sin(y), we once again apply the product rule. Let u = y and v = sin(y). The derivative of y with respect to x is dy/dx. The derivative of sin(y) with respect to x is cos(y) (dy/dx) (using the chain rule). Applying the product rule, we get: (dy/dx) sin(y) + y cos(y) (dy/dx). This example combines the product rule with the chain rule, making it a more challenging problem in implicit differentiation. We need to differentiate the product of y and sin(y), both of which are functions that implicitly depend on x. The dy/dx term appears twice in this result, once from differentiating the first y and once from differentiating sin(y). This problem reinforces the need to carefully apply both the product rule and the chain rule when dealing with implicit differentiation. It highlights the importance of recognizing when to use each rule and how to combine them effectively. The result, (dy/dx) sin(y) + y cos(y) (dy/dx), demonstrates how the derivative depends on both the original functions and the rate of change of y with respect to x. This understanding is crucial for solving more complex equations involving implicit functions. This is a great example to test one's understanding of not just the product rule, but also how to combine it with the chain rule.

i) Find dy/dx for cos(y² + 1)

To find dy/dx for cos(y² + 1), we need to apply the chain rule twice. First, the derivative of cos(u) with respect to u is -sin(u), where u = y² + 1. So, the derivative of cos(y² + 1) with respect to y² + 1 is -sin(y² + 1). Next, we need to differentiate y² + 1 with respect to x. The derivative of y² with respect to x is 2y (dy/dx), and the derivative of 1 with respect to x is 0. Therefore, the derivative of y² + 1 with respect to x is 2y (dy/dx). Applying the chain rule, we multiply these results: -sin(y² + 1) * 2y (dy/dx) = -2y sin(y² + 1) (dy/dx). This example demonstrates a more complex application of the chain rule in implicit differentiation, involving nested functions. We need to work our way through the layers of the function, differentiating each layer and multiplying the results together. The dy/dx term arises from differentiating the innermost function, y², with respect to x. This problem reinforces the importance of carefully applying the chain rule when dealing with composite functions in implicit differentiation. The result, -2y sin(y² + 1) (dy/dx), shows how the derivative depends on the sine of the composite function and the rate of change of y with respect to x. This understanding is crucial for solving problems involving complex implicit functions. It serves as a good illustration of how multiple applications of the chain rule can be required in a single problem.

j) Find dy/dx for cos(y² + x)

To find dy/dx for cos(y² + x), we again need to apply the chain rule. First, the derivative of cos(u) with respect to u is -sin(u), where u = y² + x. So, the derivative of cos(y² + x) with respect to y² + x is -sin(y² + x). Next, we need to differentiate y² + x with respect to x. The derivative of y² with respect to x is 2y (dy/dx) (using the chain rule), and the derivative of x with respect to x is 1. Therefore, the derivative of y² + x with respect to x is 2y (dy/dx) + 1. Applying the chain rule, we multiply these results: -sin(y² + x) * [2y (dy/dx) + 1] = -sin(y² + x) - 2y sin(y² + x) (dy/dx). This example demonstrates a comprehensive application of the chain rule in implicit differentiation, involving a composite function where the inner function itself is a sum of terms, one involving y and the other involving x. We need to carefully differentiate each layer of the function, remembering to apply the chain rule and the basic differentiation rules. The dy/dx term arises from differentiating the y² term with respect to x. This problem reinforces the importance of a systematic approach to implicit differentiation, carefully identifying the different layers of the function and applying the appropriate rules. The result, -sin(y² + x) - 2y sin(y² + x) (dy/dx), shows how the derivative depends on both the sine of the composite function and the rate of change of y with respect to x. This understanding is crucial for solving complex problems involving implicit functions. It serves as a solid example for demonstrating the combination of the chain rule and the sum/difference rule in implicit differentiation.

2. Implicit Differentiation with Equations

a) Find dy/dx for sin(y) + x² + 4y = cos(x)

To find dy/dx for the equation sin(y) + x² + 4y = cos(x), we need to differentiate each term with respect to x. The derivative of sin(y) with respect to x is cos(y) (dy/dx) (using the chain rule). The derivative of x² with respect to x is 2x. The derivative of 4y with respect to x is 4(dy/dx). The derivative of cos(x) with respect to x is -sin(x). Putting it all together, we get: cos(y) (dy/dx) + 2x + 4(dy/dx) = -sin(x). Now, we need to isolate dy/dx. Group the terms containing dy/dx: cos(y) (dy/dx) + 4(dy/dx) = -sin(x) - 2x. Factor out dy/dx: (dy/dx) [cos(y) + 4] = -sin(x) - 2x. Finally, solve for dy/dx: dy/dx = (-sin(x) - 2x) / (cos(y) + 4). This example demonstrates the core process of implicit differentiation: differentiating each term in the equation, applying the chain rule where necessary, and then solving for dy/dx. The steps of grouping terms, factoring, and dividing are crucial for isolating dy/dx. This problem reinforces the importance of a systematic approach to implicit differentiation, breaking down the problem into smaller, manageable steps. The result, dy/dx = (-sin(x) - 2x) / (cos(y) + 4), provides an expression for the derivative in terms of both x and y, reflecting the implicit nature of the function. This example provides a good foundation for solving more complex equations involving implicit functions. It highlights the algebraic manipulation necessary to isolate the desired derivative.

b) Find dy/dx for 3xy² + cos(y²) = 2x³ + 5

To find dy/dx for the equation 3xy² + cos(y²) = 2x³ + 5, we again differentiate each term with respect to x. For the term 3xy², we need to apply the product rule. Let u = 3x and v = y². Then, the derivative of 3x with respect to x is 3, and the derivative of y² with respect to x is 2y (dy/dx). Applying the product rule, we get: 3y² + 3x * 2y (dy/dx) = 3y² + 6xy (dy/dx). For the term cos(y²), we need to apply the chain rule twice. The derivative of cos(u) with respect to u is -sin(u), where u = y². The derivative of y² with respect to x is 2y (dy/dx). Therefore, the derivative of cos(y²) with respect to x is -sin(y²) * 2y (dy/dx) = -2y sin(y²) (dy/dx). The derivative of 2x³ with respect to x is 6x². The derivative of 5 with respect to x is 0. Putting it all together, we get: 3y² + 6xy (dy/dx) - 2y sin(y²) (dy/dx) = 6x². Now, we need to isolate dy/dx. Group the terms containing dy/dx: 6xy (dy/dx) - 2y sin(y²) (dy/dx) = 6x² - 3y². Factor out dy/dx: (dy/dx) [6xy - 2y sin(y²)] = 6x² - 3y². Finally, solve for dy/dx: dy/dx = (6x² - 3y²) / [6xy - 2y sin(y²)]. This example demonstrates a more complex application of implicit differentiation, requiring both the product rule and the chain rule. The careful application of each rule is crucial for obtaining the correct result. This problem reinforces the importance of recognizing when to apply each rule and how to combine them effectively. The result, dy/dx = (6x² - 3y²) / [6xy - 2y sin(y²)], provides an expression for the derivative in terms of both x and y, highlighting the complexity that can arise in implicit differentiation. This example provides a good challenge for students mastering the technique.

c) Find dy/dx for 5x² - x³ sin(y) + 5xy = 10

To find dy/dx for the equation 5x² - x³ sin(y) + 5xy = 10, we differentiate each term with respect to x. The derivative of 5x² with respect to x is 10x. For the term -x³ sin(y), we need to apply the product rule. Let u = -x³ and v = sin(y). The derivative of -x³ with respect to x is -3x². The derivative of sin(y) with respect to x is cos(y) (dy/dx). Applying the product rule, we get: -3x² sin(y) - x³ cos(y) (dy/dx). For the term 5xy, we also need to apply the product rule. Let u = 5x and v = y. The derivative of 5x with respect to x is 5. The derivative of y with respect to x is dy/dx. Applying the product rule, we get: 5y + 5x (dy/dx). The derivative of 10 with respect to x is 0. Putting it all together, we get: 10x - 3x² sin(y) - x³ cos(y) (dy/dx) + 5y + 5x (dy/dx) = 0. Now, we need to isolate dy/dx. Group the terms containing dy/dx: -x³ cos(y) (dy/dx) + 5x (dy/dx) = -10x + 3x² sin(y) - 5y. Factor out dy/dx: (dy/dx) [-x³ cos(y) + 5x] = -10x + 3x² sin(y) - 5y. Finally, solve for dy/dx: dy/dx = (-10x + 3x² sin(y) - 5y) / [-x³ cos(y) + 5x]. This example is another comprehensive application of implicit differentiation, requiring the product rule, the chain rule, and careful algebraic manipulation. The presence of multiple product rule applications adds complexity to the problem. This problem reinforces the need for a methodical approach to implicit differentiation, breaking down the problem into smaller steps and carefully applying the rules. The result, dy/dx = (-10x + 3x² sin(y) - 5y) / [-x³ cos(y) + 5x], provides a complex expression for the derivative, showcasing the challenges that can arise in implicit differentiation. This is an excellent example for practicing complex applications of implicit differentiation.

d) Find dy/dx for x

To find dy/dx for x, we are essentially looking for the derivative of x with respect to x. This is a basic differentiation problem. The derivative of x with respect to x is simply 1. However, the prompt asks to find dy/dx, which implies there is a function y implicitly defined in terms of x. Since the equation given is just x, there is no y term present. Thus, one interpretation is that y is a constant. If y is a constant, then dy/dx = 0. Another interpretation is that the question is ill-posed as it does not define a relationship between x and y. Without a y term, we cannot find dy/dx in the usual sense of implicit differentiation. Therefore, we can conclude that either dy/dx = 0 (if y is a constant) or the question is not well-defined. This example serves as a reminder that implicit differentiation requires a relationship between x and y. Without such a relationship, we cannot find a meaningful expression for dy/dx. The absence of a y term makes this problem an edge case in implicit differentiation, highlighting the importance of the underlying relationship between the variables. This question tests the understanding of the fundamental requirements for implicit differentiation. It is a good example to use to emphasize the importance of having a relationship between x and y to find dy/dx. This seemingly simple question emphasizes the foundational principles of implicit differentiation.