Find Dy/dx By Implicit Differentiation For Sin(x) + 5cos(5y) = 6

Jun 21, 2025 by ADMIN 65 views

Implicit Differentiation: Finding dy/dx for sin(x) + 5cos(5y) = 6

In calculus, implicit differentiation is a powerful technique used to find the derivative of a function that is not explicitly defined in the form y = f(x). Instead, implicit differentiation allows us to find dy/dx when the relationship between x and y is given implicitly, often through an equation. This method is particularly useful when it's difficult or impossible to isolate y in terms of x. Let's delve into the process of implicit differentiation and apply it to the equation sin(x) + 5cos(5y) = 6 to find dy/dx. Understanding implicit differentiation is crucial for solving a wide range of calculus problems, especially those involving related rates, inverse trigonometric functions, and equations that cannot be easily solved for one variable in terms of the other. The core idea behind implicit differentiation is to differentiate both sides of the equation with respect to x, treating y as a function of x. This requires careful application of the chain rule whenever we differentiate a term involving y. Once we have differentiated both sides, we can then algebraically solve for dy/dx. This method is applicable to a vast array of equations, including those involving trigonometric functions, as we will see in the example below. The ability to perform implicit differentiation is a fundamental skill in calculus, opening doors to solving more complex and realistic problems. Mastering this technique enables us to analyze curves and relationships that are not explicitly defined, providing a deeper understanding of the behavior of functions and their derivatives. Furthermore, implicit differentiation lays the groundwork for understanding more advanced concepts in calculus and differential equations. It allows us to handle equations where variables are intertwined, reflecting real-world scenarios where relationships are often complex and implicit rather than straightforward. This versatility makes implicit differentiation a cornerstone of calculus and a valuable tool for anyone working with mathematical models.

Understanding Implicit Differentiation

Implicit differentiation is a method used to find the derivative of a function when it is not explicitly defined as y = f(x). This technique becomes essential when dealing with equations where isolating y is either difficult or impossible. Instead of solving for y, we differentiate both sides of the equation with respect to x, treating y as a function of x. The key to implicit differentiation lies in the application of the chain rule. Whenever we differentiate a term involving y, we must multiply by dy/dx, reflecting the fact that y is dependent on x. For instance, the derivative of y^2 with respect to x is 2y(dy/dx). Similarly, the derivative of sin(y) with respect to x is cos(y)(dy/dx). The chain rule ensures that we account for the rate of change of y with respect to x when differentiating terms involving y. Once we differentiate both sides of the equation, we obtain an expression that includes dy/dx. Our next step is to algebraically manipulate this expression to isolate dy/dx on one side of the equation. This involves combining like terms, factoring out dy/dx, and dividing both sides by the appropriate expression. The resulting expression gives us dy/dx in terms of both x and y. This means that the slope of the tangent line to the curve at a given point depends on both the x-coordinate and the y-coordinate of the point. Implicit differentiation is widely used in various applications, including related rates problems, finding the derivatives of inverse functions, and analyzing curves defined by implicit equations. Its power lies in its ability to handle complex relationships between variables without the need to explicitly solve for one variable in terms of the other. This makes it an indispensable tool in calculus and related fields.

Problem Statement

Our goal is to find dy/dx for the equation sin(x) + 5cos(5y) = 6 using implicit differentiation. This equation represents an implicit relationship between x and y, meaning that y is not explicitly defined as a function of x. Instead, the relationship between x and y is given through the equation itself. To find dy/dx, we will differentiate both sides of the equation with respect to x, carefully applying the chain rule to terms involving y. The equation sin(x) + 5cos(5y) = 6 combines trigonometric functions with an implicit relationship, making it a classic example for demonstrating the power of implicit differentiation. The presence of the cosine term with a 5y argument necessitates a double application of the chain rule, adding a layer of complexity to the differentiation process. This problem highlights the importance of understanding both the basic rules of differentiation and the chain rule in the context of implicit differentiation. By working through this example, we will gain a deeper understanding of how to handle equations where variables are intertwined and how to extract the rate of change of one variable with respect to another. The result, dy/dx, will give us the slope of the tangent line to the curve defined by the equation at any point (x, y) that satisfies the equation. This information is crucial for analyzing the behavior of the curve and solving related problems in calculus and other fields. Furthermore, solving this problem reinforces the fundamental principles of implicit differentiation and prepares us for tackling more challenging problems involving implicit relationships and derivatives.

Step-by-Step Solution

Differentiate both sides of the equation with respect to x:
```
d/dx [sin(x) + 5cos(5y)] = d/dx [6]
```

The first step in solving this implicit differentiation problem is to apply the derivative operator d/dx to both sides of the given equation. This ensures that we maintain the equality while allowing us to find the relationship between the derivatives of x and y. The equation sin(x) + 5cos(5y) = 6 is the starting point, and applying the derivative to both sides is the foundation of the implicit differentiation process. We are essentially looking for how the equation changes with respect to x, considering that y is also a function of x. This initial step is crucial because it sets the stage for applying the rules of differentiation and the chain rule, which will ultimately lead us to an expression for dy/dx. The derivative of a constant, such as 6, is always zero, which simplifies the right-hand side of the equation. On the left-hand side, we have a sum of two terms, each of which requires careful application of the appropriate differentiation rules. This step highlights the fundamental principle that differentiating both sides of an equation preserves the equality and allows us to unveil the implicit relationship between the derivatives of the variables involved. It's a critical first step in unraveling the connection between x and y and finding the desired derivative, dy/dx.

Apply the derivative rules:
```
d/dx [sin(x)] + d/dx [5cos(5y)] = 0
```
We now apply the standard rules of differentiation to each term on the left-hand side of the equation. The derivative of sin(x) with respect to x is cos(x). For the second term, 5cos(5y), we need to use the chain rule because y is a function of x. The chain rule states that the derivative of a composite function f(g(x)) is f'(g(x)) * g'(x). In this case, we have a composition of functions: the outer function is 5cos(u) and the inner function is u = 5y. Applying the chain rule, we first differentiate the outer function with respect to its argument, which gives us -5sin(5y). Then, we multiply by the derivative of the inner function with respect to x, which is 5(dy/dx). Combining these results, the derivative of 5cos(5y) with respect to x is -25sin(5y)(dy/dx). This step is crucial as it transforms the original equation into one involving dy/dx, which is our target. The careful application of the chain rule is the hallmark of implicit differentiation and is essential for correctly finding the derivative when y is not explicitly defined as a function of x. This step demonstrates the power and versatility of the chain rule in handling complex derivatives and lays the groundwork for isolating dy/dx in the subsequent steps.
Differentiate each term:
```
cos(x) - 25sin(5y) * dy/dx = 0
```
This equation is obtained by differentiating each term in the previous step. The derivative of sin(x) is cos(x). The derivative of 5cos(5y) requires the chain rule. The outer function is 5cos(u), where u = 5y. The derivative of 5cos(u) with respect to u is -5sin(u), so we have -5sin(5y). Now we must multiply by the derivative of the inner function, 5y, with respect to x. The derivative of 5y with respect to x is 5(dy/dx). Combining these, we get -5sin(5y) * 5(dy/dx) = -25sin(5y)(dy/dx). The derivative of the constant 6 on the right-hand side is 0. This step is critical because it explicitly introduces the term dy/dx into the equation, which is the derivative we are trying to find. The successful application of the chain rule to the term 5cos(5y) is a key element in implicit differentiation. It allows us to account for the fact that y is a function of x, and therefore its derivative with respect to x is not simply 1, but rather dy/dx. The resulting equation now represents a relationship between cos(x), sin(5y), and dy/dx, which we can manipulate to isolate dy/dx. This step highlights the importance of careful and accurate application of derivative rules, especially the chain rule, in the context of implicit differentiation. It sets the stage for the final steps, which involve algebraic manipulation to solve for the desired derivative.
Isolate dy/dx:
```
25sin(5y) * dy/dx = cos(x)
```
To isolate dy/dx, we rearrange the equation obtained in the previous step. Our goal is to get all terms involving dy/dx on one side of the equation and all other terms on the other side. In this case, we have cos(x) - 25sin(5y)(dy/dx) = 0. To isolate the dy/dx term, we add 25sin(5y)(dy/dx) to both sides of the equation. This results in the equation cos(x) = 25sin(5y)(dy/dx). This step is a fundamental algebraic manipulation that prepares the equation for the final step of solving for dy/dx. By isolating the term containing the derivative, we are effectively separating the rate of change we are interested in from the rest of the equation. This step demonstrates the importance of algebraic skills in calculus, as the ability to manipulate equations is crucial for solving for unknown quantities. The rearrangement of terms is a common technique in many areas of mathematics and is particularly useful in calculus for solving differential equations and finding derivatives. The resulting equation now has the dy/dx term isolated on one side, making it straightforward to solve for dy/dx in the next step. This step bridges the gap between the differentiation process and the final solution, showcasing the interplay between calculus and algebra.
Solve for dy/dx:
```
dy/dx = cos(x) / (25sin(5y))
```
To finally solve for dy/dx, we divide both sides of the equation by the coefficient of dy/dx, which is 25sin(5y). Starting from the equation 25sin(5y)(dy/dx) = cos(x), we divide both sides by 25sin(5y) to isolate dy/dx on the left-hand side. This gives us the expression dy/dx = cos(x) / (25sin(5y)). This is the derivative of y with respect to x, expressed in terms of x and y. This step represents the culmination of the implicit differentiation process. We have successfully found the derivative dy/dx without having to explicitly solve the original equation for y in terms of x. The result, dy/dx = cos(x) / (25sin(5y)), provides the slope of the tangent line to the curve defined by the original equation at any point (x, y) that satisfies the equation. This expression highlights the implicit relationship between x and y and how their rates of change are related. This step underscores the power of implicit differentiation in handling equations where variables are intertwined and cannot be easily separated. The final expression for dy/dx is a valuable result that can be used to analyze the behavior of the curve, find tangent lines, and solve related problems in calculus and other fields. It demonstrates the practical application of implicit differentiation in extracting information from implicitly defined relationships.

Final Answer

Therefore, $\frac{d y}{d x} = \frac{\cos(x)}{25\sin(5y)}$ . This expression gives us the derivative of y with respect to x for the implicitly defined function sin(x) + 5cos(5y) = 6. The final answer, dy/dx = cos(x) / (25sin(5y)), encapsulates the result of the entire implicit differentiation process. It is a function of both x and y, reflecting the implicit nature of the relationship between the variables. This expression allows us to calculate the slope of the tangent line to the curve defined by the original equation at any point (x, y) that satisfies the equation. The result demonstrates the power of implicit differentiation in handling equations where y is not explicitly defined as a function of x. It provides a general formula for the derivative, which can be evaluated at specific points to obtain the slope of the tangent line at those points. This final answer is a key piece of information for analyzing the behavior of the curve and solving related problems in calculus and other fields. It represents a complete solution to the problem and showcases the effectiveness of implicit differentiation as a technique for finding derivatives of implicitly defined functions. The process involved careful application of the chain rule and algebraic manipulation to isolate dy/dx, highlighting the importance of both calculus and algebraic skills in solving such problems.