Implicit Differentiation Finding Dy/dx For Cos(xy) + X^6 = Y^6

Implicit differentiation is a powerful technique in calculus used to find the derivative of a function that is not explicitly defined in the form y = f(x). Instead, it's often used when we have an equation relating x and y, where it's difficult or impossible to isolate y. This article delves into the application of implicit differentiation to solve the equation $\cos{xy} + x^6 = y^6$, providing a step-by-step guide and a thorough explanation of the underlying concepts.

Understanding Implicit Differentiation

Before diving into the specifics of our problem, let's establish a solid understanding of implicit differentiation. In essence, implicit differentiation involves differentiating both sides of an equation with respect to x, treating y as a function of x. This is where the chain rule becomes crucial. When differentiating terms involving y, we need to multiply by dy/dx to account for the fact that y is itself a function of x. The key idea is to recognize that y is implicitly defined as a function of x, even if we don't have an explicit formula for it.

Consider a simple example: $x^2 + y^2 = 25$. This equation represents a circle centered at the origin with a radius of 5. We can't easily solve this equation for y as a single function of x (we'd get two separate functions, one for the top semicircle and one for the bottom). However, we can still find dy/dx using implicit differentiation. Differentiating both sides with respect to x, we get:

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

Now, we can solve for dy/dx:

$$\frac{dy}{dx} = -\frac{x}{y}$$

This gives us the slope of the tangent line to the circle at any point (x, y) on the circle. This example highlights the power of implicit differentiation in handling equations where y is not explicitly defined.

The beauty of implicit differentiation lies in its ability to handle complex equations where isolating y is impractical. This technique is especially useful in related rates problems, where we're interested in how the rates of change of different variables are related, and in finding derivatives of inverse functions. The underlying principle is the chain rule, which allows us to differentiate composite functions effectively. By treating y as a function of x, we can navigate the differentiation process even when an explicit formula for y is unavailable.

Applying Implicit Differentiation to $\cos{xy} + x^6 = y^6$

Now, let's tackle the given equation: $\cos{xy} + x^6 = y^6$. Our goal is to find dy/dx, which represents the rate of change of y with respect to x. This equation is a classic example where implicit differentiation shines because isolating y would be extremely difficult, if not impossible.

Step 1: Differentiate both sides with respect to x

We begin by differentiating both sides of the equation with respect to x. This is the core of implicit differentiation, where we apply the chain rule and other differentiation rules as needed. Remember, we're treating y as a function of x, so we'll need to use the chain rule whenever we differentiate a term involving y.

$$\frac{d}{dx}(\cos{xy} + x^6) = \frac{d}{dx}(y^6)$$

Step 2: Apply the chain rule and other differentiation rules

On the left side, we have the sum of two terms. We'll differentiate each term separately. For the first term, $\cos{xy}$, we need to apply the chain rule. The derivative of $\cos{u}$ is $-sin{u}$, and here, u = xy. We also need to apply the product rule to differentiate xy with respect to x. For the second term, $x^6$, we can use the power rule directly.

On the right side, we have $y^6$. Again, we need to apply the chain rule since y is a function of x. The derivative of $u^6$ is $6u^5$, and here, u = y. We'll need to multiply by dy/dx because of the chain rule.

Applying these rules, we get:

$$-\sin(xy) \frac{d}{dx}(xy) + 6x^5 = 6y^5 \frac{dy}{dx}$$

Now, we need to differentiate xy with respect to x. This requires the product rule:

$$\frac{d}{dx}(xy) = x \frac{dy}{dx} + y \frac{dx}{dx} = x \frac{dy}{dx} + y$$

Substituting this back into our equation, we have:

$$-\sin(xy) (x \frac{dy}{dx} + y) + 6x^5 = 6y^5 \frac{dy}{dx}$$

This step is crucial as it lays the foundation for solving for dy/dx. The correct application of the chain rule and product rule is paramount to obtaining the correct derivative. Double-checking each step is highly recommended to avoid errors.

Step 3: Expand and rearrange the equation

Next, we need to expand the equation and rearrange the terms to isolate dy/dx. This involves distributing the $-sin(xy)$ term and grouping all terms containing dy/dx on one side of the equation and all other terms on the other side.

Expanding the equation, we get:

$$-x \sin(xy) \frac{dy}{dx} - y \sin(xy) + 6x^5 = 6y^5 \frac{dy}{dx}$$

Now, let's move all terms containing dy/dx to the right side and the other terms to the left side:

$$6x^5 - y \sin(xy) = 6y^5 \frac{dy}{dx} + x \sin(xy) \frac{dy}{dx}$$

This rearrangement is a key step in solving for dy/dx. By grouping the terms strategically, we prepare the equation for the final isolation of the derivative.

Step 4: Factor out dy/dx

Now, we factor out dy/dx from the terms on the right side of the equation. This is a straightforward step that simplifies the equation and allows us to isolate dy/dx in the next step.

Factoring out dy/dx, we get:

$$6x^5 - y \sin(xy) = (6y^5 + x \sin(xy)) \frac{dy}{dx}$$

This step is a crucial algebraic manipulation that brings us closer to our goal of finding dy/dx. Factoring out the derivative allows us to treat the expression in parentheses as a coefficient, which can then be divided to isolate dy/dx.

Step 5: Solve for dy/dx

Finally, we solve for dy/dx by dividing both sides of the equation by the expression in parentheses. This gives us the derivative dy/dx in terms of x and y.

Dividing both sides by $(6y^5 + x \sin(xy))$, we get:

$$\frac{dy}{dx} = \frac{6x^5 - y \sin(xy)}{6y^5 + x \sin(xy)}$$

This is the final result. We have successfully found dy/dx using implicit differentiation. The derivative is expressed in terms of both x and y, which is typical in implicit differentiation problems. This result gives us the slope of the tangent line to the curve defined by the original equation at any point (x, y) that satisfies the equation.

Conclusion

In this article, we have demonstrated the application of implicit differentiation to find dy/dx for the equation $\cos{xy} + x^6 = y^6$. We've walked through the process step-by-step, emphasizing the importance of the chain rule and product rule in this context. Implicit differentiation is a powerful tool for finding derivatives when y is not explicitly defined as a function of x. This technique is widely used in calculus and has applications in various fields, including physics, engineering, and economics. By mastering implicit differentiation, you gain a valuable skill for tackling complex differentiation problems.

This step-by-step approach provides a clear roadmap for solving similar problems. The key takeaways are the correct application of the chain rule, the product rule, and the algebraic manipulations required to isolate dy/dx. Implicit differentiation is not just a technique; it's a way of thinking about derivatives when the relationship between variables is not explicitly defined. This skill is invaluable for anyone pursuing further studies in mathematics or related fields.

The result, $\frac{dy}{dx} = \frac{6x^5 - y \sin(xy)}{6y^5 + x \sin(xy)}$, represents the slope of the tangent line to the curve defined by the original equation at any point (x, y) that satisfies the equation. This result can be used to analyze the behavior of the curve, find critical points, and solve related rates problems. The power of implicit differentiation lies in its ability to provide this information even when an explicit formula for y in terms of x is unavailable. This makes it an indispensable tool in calculus and its applications.