Tangent Line Equation For Y=7ln((e^x+e^-x)/2) At (0,0)

Aug 7, 2025 by ADMIN 55 views

Hey everyone! Today, we're diving into a super interesting calculus problem: finding the equation of the tangent line to the graph of the function y = 7ln((e^x + e^-x) / 2) at the point (0, 0). This problem combines concepts from exponential functions, logarithms, and derivatives, so it's a fantastic way to flex our calculus muscles. Let's break it down step-by-step!

Understanding the Problem

Before we jump into the math, let's make sure we understand what we're trying to do. We have a function, y = 7ln((e^x + e^-x) / 2), which looks a bit intimidating, but we'll tackle it. We want to find the equation of a line that touches the graph of this function at only one point – the point (0, 0). This special line is called the tangent line. Think of it like a straight line that just barely grazes the curve at that specific spot. The key to finding this tangent line is to figure out its slope at the point (0, 0). And how do we find the slope? That's right, derivatives!

In essence, the problem requires us to apply our knowledge of differential calculus to determine the equation of a line that is tangent to a given curve at a specific point. To achieve this, we'll need to find the derivative of the function, which will give us the slope of the tangent line at any point on the curve. Then, we'll evaluate the derivative at the given point (0, 0) to find the slope of the tangent line at that specific location. Finally, using the point-slope form of a linear equation, we'll construct the equation of the tangent line. It might sound like a lot, but we'll take it one step at a time, making sure everything is clear along the way. Remember, the beauty of calculus lies in its ability to solve complex problems by breaking them down into smaller, manageable parts.

Step 1: Finding the Derivative

The first step in finding the tangent line is to calculate the derivative of the function. This is where our knowledge of calculus comes in handy. The function we're working with is y = 7ln((e^x + e^-x) / 2). Notice that this is a composition of functions, so we'll need to use the chain rule. Remember the chain rule? It states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. This is a fundamental concept in calculus, and it's crucial for tackling problems like this one.

Let's break down the function into its components. We have an outer function, which is 7ln(u), where u is the inner function. The inner function is (e^x + e^-x) / 2. So, let's first find the derivative of the outer function with respect to u. The derivative of 7ln(u) is simply 7/u. Now, we need to find the derivative of the inner function, (e^x + e^-x) / 2, with respect to x. This involves differentiating exponential functions, which we know are their own derivatives (with a slight tweak for e^-x). The derivative of e^x is just e^x, and the derivative of e^-x is -e^-x (don't forget the negative sign from the chain rule!). So, the derivative of (e^x + e^-x) / 2 is (e^x - e^-x) / 2. Now, we can apply the chain rule. We multiply the derivative of the outer function (7/u) by the derivative of the inner function ((e^x - e^-x) / 2). This gives us:

dy/dx = 7 / ((e^x + e^-x) / 2) * (e^x - e^-x) / 2

We can simplify this a bit by canceling out the 2s and rearranging terms:

dy/dx = 7(e^x - e^-x) / (e^x + e^-x)

This is the derivative of our original function, and it represents the slope of the tangent line at any point x on the curve. Now we're one step closer to finding the equation of the tangent line at (0, 0).

Step 2: Finding the Slope at (0,0)

Now that we have the derivative, dy/dx = 7(e^x - e^-x) / (e^x + e^-x), we can find the slope of the tangent line at the point (0, 0). Remember, the derivative gives us the slope of the tangent line at any point x. So, to find the slope at (0, 0), we simply need to plug in x = 0 into our derivative. This is a crucial step because it gives us the specific slope we need to define our tangent line. It’s like zooming in on the curve at the point (0, 0) and finding the exact steepness of the line that touches the curve there.

Let's do it:

dy/dx |_(x=0) = 7(e^0 - e^-0) / (e^0 + e^-0)

Remember that any number raised to the power of 0 is 1. So, e^0 = 1 and e^-0 = 1. Substituting these values into our equation, we get:

dy/dx |_(x=0) = 7(1 - 1) / (1 + 1)

This simplifies to:

dy/dx |_(x=0) = 7(0) / 2 = 0

So, the slope of the tangent line at the point (0, 0) is 0. This means the tangent line is horizontal at this point. This makes sense if you think about the shape of the function – it likely has a minimum or a point of inflection at x = 0. A horizontal tangent line indicates that the function is momentarily neither increasing nor decreasing at that point. Now that we have the slope, we're ready to find the equation of the tangent line.

Step 3: Finding the Equation of the Tangent Line

We're in the home stretch! We know the slope of the tangent line at (0, 0) is 0, and we know the tangent line passes through the point (0, 0). Now we just need to use this information to find the equation of the line. There are a couple of ways to do this, but the most straightforward is using the point-slope form of a linear equation. Remember the point-slope form? It's a handy formula that lets us write the equation of a line if we know its slope and a point it passes through. Mastering different forms of linear equations is a key skill in algebra and calculus, and the point-slope form is particularly useful in situations like this.

The point-slope form is given by:

y - y1 = m(x - x1)

where:

m is the slope of the line
(x1, y1) is a point on the line

In our case, m = 0 (the slope we calculated) and (x1, y1) = (0, 0) (the given point). Plugging these values into the point-slope form, we get:

y - 0 = 0(x - 0)

This simplifies to:

y = 0

Therefore, the equation of the tangent line to the graph of the function y = 7ln((e^x + e^-x) / 2) at the point (0, 0) is y = 0. This is a horizontal line that coincides with the x-axis. It makes sense that the tangent line is horizontal because we found the slope to be 0. This result highlights the power of calculus in finding geometric properties of curves.

Conclusion

So, there you have it! We successfully found the equation of the tangent line to the given function at the specified point. We started by understanding the problem and identifying the key concepts involved. Then, we systematically worked through the steps: finding the derivative using the chain rule, evaluating the derivative at the given point to find the slope, and finally, using the point-slope form to write the equation of the tangent line. This problem demonstrates the interconnectedness of different calculus concepts and how they can be applied to solve real-world problems.

This was a fun problem that let us practice a bunch of calculus skills. We tackled the chain rule, exponential functions, and logarithmic functions all in one go! Remember, the key to mastering calculus is practice. The more problems you solve, the more comfortable you'll become with the techniques and the better you'll understand the underlying concepts. So keep practicing, keep exploring, and keep having fun with math! The answer is y = 0.