Finding Y'(9) Using Implicit Differentiation Given Sqrt(x) + Sqrt(y) = 12 And Y(9) = 81

Hey guys! Let's dive into a cool math problem where we need to find the derivative $y'(9)$ using implicit differentiation. This might sound intimidating, but trust me, we'll break it down step by step. We're given the equation $\sqrt{x} + \sqrt{y} = 12$ and the condition $y(9) = 81$. Our mission is to find the value of $y'$ when $x = 9$. So, grab your thinking caps, and let's get started!

Understanding the Problem

Before we jump into solving, let's make sure we understand what's going on. We have an equation that relates $x$ and $y$, but it's not in the typical $y = f(x)$ form. This is where implicit differentiation comes in handy. Implicit differentiation is a technique we use when we can't easily isolate $y$ in terms of $x$. The key idea is to differentiate both sides of the equation with respect to $x$, treating $y$ as a function of $x$. Remember the chain rule? It's going to be our best friend here!

We're also given the condition $y(9) = 81$. This tells us that when $x = 9$, $y = 81$. This is crucial information because we'll need these values to find the specific value of $y'(9)$. Think of $y'(9)$ as the slope of the tangent line to the curve defined by our equation at the point $(9, 81)$. In essence, the goal here is to calculate that slope using the given information and our implicit differentiation skills.

Step-by-Step Solution

Okay, let’s get our hands dirty with some math! Here’s how we’ll tackle this problem:

1. Implicit Differentiation

First, we'll differentiate both sides of the equation $\sqrt{x} + \sqrt{y} = 12$ with respect to $x$. Remember, we'll treat $y$ as a function of $x$, so we'll need to use the chain rule when differentiating terms involving $y$.

The equation is:

$\sqrt{x} + \sqrt{y} = 12$

Rewrite the square roots as exponents:

$x^{1/2} + y^{1/2} = 12$

Now, differentiate both sides with respect to $x$:

$\frac{d}{dx}(x^{1/2} + y^{1/2}) = \frac{d}{dx}(12)$

Applying the power rule and the chain rule, we get:

$\frac{1}{2}x^{-1/2} + \frac{1}{2}y^{-1/2} \frac{dy}{dx} = 0$

Notice the $\frac{dy}{dx}$ term? That's $y'$, which is what we're trying to find! This step is the core of implicit differentiation, where we handle the derivative of $y$ with respect to $x$.

2. Isolate y'

Next, we need to isolate $y'$ (which is $\frac{dy}{dx}$) in the equation we just obtained. Let's rearrange the terms:

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

Multiply both sides by 2 to get rid of the fractions:

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

Now, divide both sides by $y^{-1/2}$ to isolate $\frac{dy}{dx}$:

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

We can simplify this expression by rewriting the negative exponents:

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

Or, even more simply:

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

Great! We now have an expression for $y'$ in terms of $x$ and $y$. Isolating $y'$ is a crucial algebraic step that allows us to find the derivative's value at a specific point.

3. Use the Given Condition y(9) = 81

Remember the condition $y(9) = 81$? This is where it comes into play. We know that when $x = 9$, $y = 81$. We'll plug these values into the expression we found for $y'$:

$y'(9) = -\sqrt{\frac{81}{9}}$

4. Calculate y'(9)

Now, let's do the math:

$y'(9) = -\sqrt{9}$

$y'(9) = -3$

And there we have it! We've found that $y'(9) = -3$. This is the final numerical answer that gives us the slope of the tangent line at the specified point.

Conclusion

So, guys, we successfully found $y'(9)$ using implicit differentiation! We started with the equation $\sqrt{x} + \sqrt{y} = 12$ and the condition $y(9) = 81$. We differentiated implicitly, isolated $y'$, plugged in the given values, and calculated the result. The value of $y'(9)$ is -3.

To recap, the key steps were:

1. Implicitly differentiate the given equation with respect to $x$.
2. Isolate $y'$ in the resulting equation.
3. Substitute the given values of $x$ and $y$ (from the condition $y(9) = 81$).
4. Calculate the value of $y'$.

Implicit differentiation is a powerful tool in calculus, and this problem nicely illustrates how it works. Practice makes perfect, so try tackling similar problems to solidify your understanding. Keep up the great work, and happy calculating!

Additional Practice Problems

To further enhance your understanding of implicit differentiation, try solving these practice problems:

1. Given $x^2 + y^2 = 25$ and $y(3) = 4$, find $y'(3)$.
2. Given $x^3 + y^3 = 6xy$, find $\frac{dy}{dx}$.
3. Given $sin(x) + cos(y) = 1$ and $y(\frac{\pi}{2}) = 0$, find $y'(\frac{\pi}{2})$.

Working through these problems will help you become more comfortable with the process and variations of implicit differentiation. Remember to focus on applying the chain rule correctly and isolating $y'$ effectively.

Common Mistakes to Avoid

While working with implicit differentiation, it's easy to make common mistakes. Here are a few to keep in mind:

• Forgetting the Chain Rule: When differentiating terms involving $y$, always remember to multiply by $\frac{dy}{dx}$ (or $y'$). This is the most frequent error.
• Incorrect Differentiation: Make sure you know the basic differentiation rules (power rule, trigonometric derivatives, etc.) and apply them correctly.
• Algebraic Errors: Mistakes during the isolation of $y'$ can lead to incorrect results. Double-check each step of your algebraic manipulations.
• Substituting Too Early: Avoid substituting the given values for $x$ and $y$ before isolating $y'$. This can complicate the process.

By being aware of these common pitfalls, you can significantly improve your accuracy when using implicit differentiation.

Importance of Implicit Differentiation

Implicit differentiation isn't just a mathematical exercise; it has practical applications in various fields. Here are a few examples:

• Related Rates Problems: Implicit differentiation is crucial in problems where we need to find the rate of change of one variable with respect to time, given the rate of change of another related variable. Think of scenarios like the changing volume of a balloon or the distance between two moving objects.
• Curve Sketching: The derivative $y'$ obtained through implicit differentiation helps us analyze the slope of a curve at different points, which is essential for sketching the curve.
• Optimization Problems: In some optimization problems, the relationship between variables is implicitly defined. Implicit differentiation helps us find critical points and determine maximum or minimum values.
• Physics and Engineering: Many physical laws and engineering principles are expressed as implicit equations. Implicit differentiation is a valuable tool for analyzing these relationships.

Understanding implicit differentiation expands your problem-solving toolkit and enables you to tackle a wider range of mathematical and real-world challenges. Keep practicing, and you'll find it becomes a natural and powerful technique in your mathematical arsenal.