Hey guys, let's dive into a cool math problem! We're gonna figure out the derivative (dy/dx) of a function, but with a bit of a twist. The function is defined implicitly, meaning it's not directly in the form of y = something with x. Instead, we have x to the power of y equals the natural logarithm of x, or in math speak, x^y = log(x). And the kicker? We need to find the value of this derivative at a specific point, when x is equal to e (Euler's number, approximately 2.718).

This type of problem is super important in calculus, because it forces us to use a technique called implicit differentiation. We can't just rearrange the equation to get y by itself, so we have to treat y as a function of x and use the chain rule. Sounds a little intimidating, right? Don't sweat it. We'll break it down step by step so you can totally nail it. We will use the concepts of implicit differentiation, logarithmic differentiation, and the chain rule. We will be taking a look at each of them individually in the following paragraphs, so stick around and pay close attention to each section of the article.

Understanding Implicit Differentiation and Logarithmic Differentiation

Alright, let's get our heads around implicit differentiation. Basically, it's a technique we use when we can't easily isolate y in an equation. Instead of trying to solve for y, we differentiate both sides of the equation with respect to x, treating y as a function of x. Whenever we differentiate a term containing y, we have to remember to multiply by dy/dx because of the chain rule. Think of it like this: if you have something like y^2, its derivative is 2y * (dy/dx). This is because when we take the derivative of y with respect to x, we have to account for how y changes with respect to x, hence the dy/dx. That little dy/dx is the key to implicit differentiation.

Now, let's chat about logarithmic differentiation. This is our secret weapon when we have complicated expressions involving exponents, products, and quotients. The main idea is to take the natural logarithm of both sides of the equation. This does two amazing things: it simplifies exponents (using the power rule of logarithms) and it often makes the equation much easier to differentiate. Logarithmic differentiation is particularly useful when the variable is in both the base and the exponent, like we have in x^y.

So, how does this relate to our problem? Well, our equation, x^y = log(x), is perfect for using logarithmic differentiation. The x^y has both a variable base and a variable exponent. This is where logarithmic differentiation becomes our best friend. By taking the natural log of both sides, we can simplify the equation and make it much easier to differentiate implicitly. This will allow us to isolate dy/dx and solve for its value at the given point.

We'll use these two techniques to unravel the problem. We start by taking the natural logarithm of both sides to simplify the equation, then differentiate implicitly, and finally, we solve for dy/dx. It might seem a bit complicated at first, but with practice, it's really not that bad. Implicit differentiation and logarithmic differentiation are fundamental tools in calculus, and understanding them opens doors to solving a wide range of interesting problems. So let's keep going and see how it all comes together! Trust me; it's going to be a rewarding journey. These concepts are used in many real-world applications, such as optimization problems, related rates problems, and analyzing the behavior of functions.

Step-by-step: Differentiating x^y = log(x)

Okay, guys, let's get down to the nitty-gritty and work through the problem step by step. We're starting with our equation: x^y = log(x). Our goal is to find dy/dx, the derivative of y with respect to x.

First, we're going to use logarithmic differentiation. As mentioned, taking the natural logarithm of both sides of the equation is the first step. This gives us:

log(x^y) = log(log(x))

Using the power rule of logarithms (which states that log(a^b) = b * log(a)), we can simplify the left side:

y * log(x) = log(log(x))

Now we've got a much friendlier equation to work with. The exponents are gone, and we have a product on the left side. It's now time for the next step: using implicit differentiation. We'll differentiate both sides of the equation with respect to x. Remember, whenever we differentiate a term with y, we need to multiply by dy/dx.

Let's break it down term by term. On the left side, we have a product y * log(x). We need to use the product rule, which states that the derivative of u*v is u'v + uv'. So, the derivative of y * log(x) is (dy/dx) * log(x) + y * (1/x).

On the right side, we have log(log(x)). This requires the chain rule. The derivative of log(u) is (1/u) * u'. Therefore, the derivative of log(log(x)) is (1/log(x)) * (1/x). Putting it all together, our differentiated equation becomes:

(dy/dx) * log(x) + y/x = (1/x) / log(x)

Next, we need to solve for dy/dx. Let's isolate the dy/dx term. First, subtract y/x from both sides:

(dy/dx) * log(x) = (1/x) / log(x) - y/x

Then, divide both sides by log(x):

dy/dx = [(1/x) / log(x) - y/x] / log(x)

This is the general expression for dy/dx. However, we're not done yet. We need to find the value of dy/dx at x = e. To do that, we need to find the value of y when x = e. This is the point where the whole thing comes together. By using these concepts, you can solve similar problems, gaining a better understanding of calculus.

Finding the value of y at x = e

Alright, before we can find dy/dx at x = e, we need to find the value of y when x = e. Remember our original equation: x^y = log(x). We want to find the value of y when x = e. Let's substitute x = e into the original equation:

e^y = log(e)

Since log(e) is equal to 1 (because the natural logarithm of e is 1), the equation simplifies to:

e^y = 1

To solve for y, we need to figure out what power we need to raise e to get 1. The answer, of course, is 0. So, when x = e, y = 0. Therefore, the point we are interested in is (e, 0). This step is super important, because the value of y will be needed when calculating the final answer. We have to make sure to calculate it first to avoid any confusion or mistake during the evaluation of the derivative. Now we have everything we need to calculate the value of the derivative.

Calculating dy/dx at x = e

Now for the grand finale! We're finally ready to plug in x = e and y = 0 into our equation for dy/dx and find the final answer. Remember our equation for dy/dx from before:

dy/dx = [(1/x) / log(x) - y/x] / log(x)

Let's substitute x = e and y = 0:

dy/dx = [(1/e) / log(e) - 0/e] / log(e)

We know that log(e) = 1, so the equation becomes:

dy/dx = [(1/e) / 1 - 0] / 1

This simplifies to:

dy/dx = 1/e

And there you have it! The value of dy/dx at x = e is 1/e. This means that at the point (e, 0) on the curve defined by x^y = log(x), the slope of the tangent line is 1/e. Congratulations, guys, you've successfully navigated a tricky calculus problem using implicit and logarithmic differentiation. The result is a number slightly less than 0.4. This also means that, at the point where x is e, the function is increasing in value. Awesome, right? This entire process of solving the problem can be used to solve more complex problems with different functions, so practice this method and you will be ready for the next problem. Well done!

Summary

Let's recap what we've done:

• Started with the equation x^y = log(x).
• Used logarithmic differentiation to simplify it.
• Differentiated implicitly to find an expression for dy/dx.
• Found the value of y when x = e.
• Substituted x = e and y = 0 into the dy/dx equation.
• Calculated the final answer: dy/dx = 1/e at x = e.

This problem nicely illustrates the power of implicit differentiation and logarithmic differentiation. These are essential tools in calculus that allow us to deal with complex and non-standard functions. Keep practicing, and you'll become a pro at these types of problems! You can easily adapt this method to solve any problems that come your way, so keep practicing and you will do well.

I hope you enjoyed this journey. Keep up the math, and I will see you in the next one! Don't forget to practice and apply what you learned today, and you will see the results!