Find Y'(3) Using Implicit Differentiation: A Step-by-Step Guide

Hey guys! Today, we're diving into a classic calculus problem: finding the derivative $y'(3)$ using implicit differentiation. This is a super useful technique when you can't easily solve for $y$ explicitly in terms of $x$. We'll break down each step, making it easy to follow along. So, grab your pencils, and let's get started!

The Problem

We're given the equation:

$2x^2 + 4x + xy = 3$

and the condition $y(3) = -9$. Our mission, should we choose to accept it (and we do!), is to find $y'(3)$.

Step 1: Implicit Differentiation

The name of the game is implicit differentiation. We're going to differentiate both sides of the equation with respect to $x$, keeping in mind that $y$ is a function of $x$, i.e., $y = y(x)$. This means we'll need to use the chain rule when differentiating terms involving $y$.

Differentiating both sides of $2x^2 + 4x + xy = 3$ with respect to $x$ gives us:

$\frac{d}{dx}(2x^2) + \frac{d}{dx}(4x) + \frac{d}{dx}(xy) = \frac{d}{dx}(3)$

Let's break down each term:

• $\frac{d}{dx}(2x^2) = 4x$
• $\frac{d}{dx}(4x) = 4$
• $\frac{d}{dx}(xy)$: Here, we need the product rule: $\frac{d}{dx}(uv) = u'v + uv'$. So, $\frac{d}{dx}(xy) = (1)(y) + (x)\frac{dy}{dx} = y + x\frac{dy}{dx}$
• $\frac{d}{dx}(3) = 0$ (since the derivative of a constant is zero)

Putting it all together, we have:

$4x + 4 + y + x\frac{dy}{dx} = 0$

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

Our goal now is to isolate $\frac{dy}{dx}$, which is the same as $y'$. Let's rearrange the equation:

$x\frac{dy}{dx} = -4x - 4 - y$

Now, divide by $x$:

$\frac{dy}{dx} = \frac{-4x - 4 - y}{x}$

So, we have found an expression for $y'$ in terms of $x$ and $y$:

$y' = \frac{-4x - 4 - y}{x}$

This formula encapsulates the rate of change of y with respect to x for any point (x, y) that satisfies the original equation. It's a powerful result of implicit differentiation.

Step 3: Evaluate $y'(3)$

We want to find $y'(3)$, which means we need to plug in $x = 3$. We also know that $y(3) = -9$. Substituting these values into our expression for $y'$ gives us:

$y'(3) = \frac{-4(3) - 4 - (-9)}{3}$

$y'(3) = \frac{-12 - 4 + 9}{3}$

$y'(3) = \frac{-7}{3}$

So, $y'(3) = -\frac{7}{3}$.

Conclusion

And there you have it! We successfully found $y'(3)$ using implicit differentiation. Remember, the key steps are:

1. Differentiate implicitly: Differentiate both sides of the equation with respect to $x$, using the chain rule where necessary.
2. Solve for $\frac{dy}{dx}$: Isolate $y'$ in the resulting equation.
3. Evaluate: Plug in the given values of $x$ and $y$ to find the specific value of $y'$.

Implicit differentiation might seem tricky at first, but with practice, you'll become a pro. Keep practicing, and you'll be solving these problems in your sleep! This method is invaluable when dealing with equations where y cannot be easily isolated, opening doors to solving a wider array of calculus problems. Remember to always double-check your work and ensure you're applying the chain rule and product rule correctly. Understanding these fundamental principles will solidify your grasp on implicit differentiation and its applications. Keep exploring, keep learning, and most importantly, keep having fun with math! Also, remember the importance of the product rule when differentiating terms that involve products of x and y. A common mistake is to forget to apply the product rule, leading to incorrect results. This method is used not only in math but can be applied to economics and physics as well.

More Examples and Tips

Let's solidify your understanding with another quick example. Consider the equation $x^2 + y^2 = 25$. We want to find $y'$ using implicit differentiation.

Step 1: Differentiate implicitly

$\frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}(25)$

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

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

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

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

So, $y' = -\frac{x}{y}$.

Key Tips for Success:

• Practice, practice, practice: The more you work through problems, the more comfortable you'll become with the process.
• Pay attention to the chain rule: This is crucial when differentiating terms involving $y$.
• Double-check your algebra: Make sure you're solving for $\frac{dy}{dx}$ correctly.
• Understand the question: Know exactly what you're being asked to find.

By following these tips and working through examples, you'll master implicit differentiation and be able to tackle even the trickiest calculus problems. Always remember to review the fundamental rules of differentiation, such as the power rule, product rule, quotient rule, and chain rule. These rules are the building blocks of calculus, and a strong understanding of them will greatly enhance your ability to solve complex problems. Don't be afraid to seek help from your instructor or classmates if you're struggling with any concepts. Collaboration and discussion can often lead to a deeper understanding of the material. Additionally, consider using online resources such as Khan Academy or Paul's Online Math Notes for extra practice and explanations. These resources offer a wealth of information and can be invaluable tools in your learning journey. Keep challenging yourself with new problems and exploring different applications of implicit differentiation. The more you engage with the material, the more confident and proficient you'll become. Remember, learning calculus is a journey, not a destination, so enjoy the process and celebrate your progress along the way. Also remember that $y'$ is just a notation for the derivative of the function $y(x)$.

Implicit differentiation is not just a theoretical concept; it has numerous real-world applications in various fields. For example, in physics, it is used to analyze the motion of objects along curved paths. In economics, it helps to determine the rates of change of related variables, such as supply and demand. In computer graphics, it is used to render smooth curves and surfaces. By understanding the principles of implicit differentiation, you can gain a deeper appreciation for its practical significance and its role in solving real-world problems. As you continue your studies in mathematics and related fields, you will encounter many more applications of this powerful technique. Keep exploring, keep questioning, and keep discovering the beauty and utility of mathematics.

A Final Word

So there you have it, a comprehensive guide to solving implicit differentiation problems. Keep practicing, and you'll be differentiating like a pro in no time! Happy calculating! Remember the value of online calculators to verify your answer.