Finding Dy/dx: A Step-by-Step Solution

Hey everyone! Let's dive into a common problem in calculus: finding the derivative ${\frac{dy}{dx}}$. In this article, we'll break down a problem where we're given an equation relating ${x}$ and ${y}$, specifically ${x^3y - 8 = y + 5}$, and we need to find ${\frac{dy}{dx}}$, assuming ${y = f(x)}$ is a differentiable function. This involves a technique called implicit differentiation, and we'll go through it step by step so you can master it too!

Understanding Implicit Differentiation

Before we jump into solving the problem, let's quickly recap what implicit differentiation is all about. Sometimes, instead of having ${y}$ explicitly defined as a function of ${x}$ (like ${y = x^2 + 3}$), we have an equation where ${x}$ and ${y}$ are mixed together. Our given equation, ${x^3y - 8 = y + 5}$, is a perfect example of this. Implicit differentiation allows us to find ${\frac{dy}{dx}}$ even when we can't easily isolate ${y}$. The key idea is to differentiate both sides of the equation with respect to ${x}$, remembering that ${y}$ is a function of ${x}$, so we'll need to use the chain rule when differentiating terms involving ${y}$. This might sound complicated, but it’s totally manageable once you get the hang of it!

Let's talk about the underlying principles in more detail. When we differentiate a term involving ${y}$ with respect to ${x}$, we're essentially asking how ${y}$ changes as ${x}$ changes. Since ${y}$ is a function of ${x}$, a change in ${x}$ will generally lead to a change in ${y}$. The chain rule helps us capture this relationship. For instance, if we have a term like ${y^2}$, differentiating it with respect to ${x}$ gives us ${2y \frac{dy}{dx}}$. We first differentiate ${y^2}$ with respect to ${y}$ (which is ${2y}$), and then we multiply by ${\frac{dy}{dx}}$ to account for the fact that ${y}$ is changing with ${x}$. Similarly, when we encounter terms like ${x^3y}$, we need to use the product rule in addition to the chain rule. This is because we have a product of two functions of ${x}$, namely ${x^3}$ and ${y}$. The product rule tells us that the derivative of ${uv}$ (where ${u}$ and ${v}$ are functions of ${x}$) is ${u'v + uv'}$. So, the derivative of ${x^3y}$ will be ${3x^2y + x^3 \frac{dy}{dx}}$. Understanding these core concepts is crucial for successfully tackling implicit differentiation problems. Now that we have a solid grasp of the theory, let’s apply it to our specific problem.

Step-by-Step Solution

Okay, let's tackle the problem head-on! We're given the equation ${x^3y - 8 = y + 5}$, and our mission is to find ${\frac{dy}{dx}}$. Here's how we'll do it, step by step:

1. Differentiate both sides with respect to x

This is the heart of implicit differentiation. We apply the derivative operator ${\frac{d}{dx}}$ to both sides of the equation. Remember, what we do to one side, we must do to the other to maintain equality. So, we get:

${\frac{d}{dx}(x^3y - 8) = \frac{d}{dx}(y + 5)}$

2. Apply the product rule and chain rule

On the left side, we have ${x^3y}$, which is a product of two functions of ${x}$. We'll need the product rule here. The derivative of ${x^3}$ is ${3x^2}$, and the derivative of ${y}$ with respect to ${x}$ is ${\frac{dy}{dx}}$. So, applying the product rule, we get:

${\frac{d}{dx}(x^3y) = 3x^2y + x^3\frac{dy}{dx}}$

The derivative of the constant ${-8}$ is simply 0. On the right side, the derivative of ${y}$ with respect to ${x}$ is ${\frac{dy}{dx}}$, and the derivative of the constant 5 is 0. So, putting it all together, our equation becomes:

${3x^2y + x^3\frac{dy}{dx} - 0 = \frac{dy}{dx} + 0}$

Which simplifies to:

${3x^2y + x^3\frac{dy}{dx} = \frac{dy}{dx}}$

3. Collect terms with dy/dx

Our goal is to isolate ${\frac{dy}{dx}}$. So, let's gather all the terms containing ${\frac{dy}{dx}}$ on one side of the equation and all the other terms on the other side. Subtracting ${\frac{dy}{dx}}$ from both sides gives us:

${x^3\frac{dy}{dx} - \frac{dy}{dx} = -3x^2y}$

4. Factor out dy/dx

Now, we can factor out ${\frac{dy}{dx}}$ from the left side:

${\frac{dy}{dx}(x^3 - 1) = -3x^2y}$

5. Solve for dy/dx

Finally, to get ${\frac{dy}{dx}}$ by itself, we divide both sides by ${(x^3 - 1)}$:

${\frac{dy}{dx} = \frac{-3x^2y}{x^3 - 1}}$

And there you have it! We've found ${\frac{dy}{dx}}$ using implicit differentiation. This is our general solution for the derivative. If we had specific values for ${x}$ and ${y}$, we could plug them in to find the value of the derivative at that point.

Common Mistakes to Avoid

Implicit differentiation can be tricky, and it's easy to make mistakes if you're not careful. Let's go over some common pitfalls to watch out for:

• Forgetting the chain rule: This is probably the most common mistake. Remember, when you differentiate a term involving ${y}$ with respect to ${x}$, you need to multiply by ${\frac{dy}{dx}}$. For example, the derivative of ${y^2}$ with respect to ${x}$ is ${2y\frac{dy}{dx}}$, not just ${2y}$.
• Misapplying the product rule: When you have a product of two functions of ${x}$ (like ${x^3y}$), you must use the product rule. Don't forget to differentiate both parts of the product and add them together correctly. The derivative of ${x^3y}$ is ${3x^2y + x^3\frac{dy}{dx}}$, not just ${3x^2y}$ or ${x^3\frac{dy}{dx}}$.
• Algebra errors: After you've differentiated, you'll need to manipulate the equation to isolate ${\frac{dy}{dx}}$. This involves algebraic steps like collecting terms, factoring, and dividing. Be careful with your algebra to avoid making mistakes that will lead to an incorrect answer.
• Not differentiating constants correctly: The derivative of a constant is always 0. Make sure you don't forget this when differentiating equations involving constants.
• Rushing through the steps: Implicit differentiation requires careful attention to detail. It's easy to make a mistake if you rush. Take your time, write out each step clearly, and double-check your work.

By being aware of these common mistakes, you can significantly improve your accuracy and confidence when working with implicit differentiation.

Practice Problems

To really master implicit differentiation, you need to practice! Here are a few problems you can try on your own:

1. Find ${\frac{dy}{dx}}$ if ${x^2 + y^2 = 25}$.
2. Find ${\frac{dy}{dx}}$ if ${x^2y + xy^2 = 3x}$.
3. Find ${\frac{dy}{dx}}$ if ${\sin(xy) = x^2 + 1}$.

Working through these problems will help you solidify your understanding of the process and identify any areas where you might need more practice. Remember to follow the steps we outlined earlier: differentiate both sides, apply the product and chain rules as needed, collect terms with ${\frac{dy}{dx}}$, factor, and solve for ${\frac{dy}{dx}}$. Don't be afraid to make mistakes – that's how we learn! And if you get stuck, revisit the examples and explanations we've discussed.

Conclusion

So there you have it! We've walked through how to find ${\frac{dy}{dx}}$ using implicit differentiation with a detailed example. Remember the key steps: differentiate both sides, apply the product and chain rules, collect ${\frac{dy}{dx}}$ terms, factor, and solve. With practice, you'll become a pro at implicit differentiation. Keep practicing, and you'll be solving these problems like a champ in no time! Keep up the great work, guys! You've got this! And always remember, the more you practice, the easier it gets. So, keep those pencils moving and those brains working, and you'll be mastering calculus concepts in no time. Happy differentiating!