Tangent Line Slope & Derivative Of 1/x: Step-by-Step

Hey guys! Let's tackle some calculus problems today. We're going to dive into finding the slope of a tangent line and calculating derivatives. Specifically, we'll work through an example involving the function f(x) = (x-1)/(x+1) and then find the derivative of f(x) = 1/x. So, grab your pencils, and let's get started!

Finding the Slope of the Tangent Line to f(x) = (x-1)/(x+1) at x = 0

The big question here is: how do we find the slope of the tangent line to the function f(x) = (x-1)/(x+1) at the point x = 0? This is a classic calculus problem that combines the concept of derivatives with the geometric interpretation of a tangent line. Remember, the slope of the tangent line at a specific point on a curve is given by the derivative of the function evaluated at that point. So, our mission is clear: first, we need to find the derivative of f(x), and then we'll plug in x = 0 to get our answer. But how exactly do we find the derivative of a function that looks like a fraction? That's where the quotient rule comes into play. The quotient rule is a fundamental tool in calculus that allows us to differentiate functions that are expressed as the ratio of two other functions. It states that if we have a function f(x) = u(x) / v(x), then its derivative f'(x) is given by: f'(x) = [u'(x)v(x) - u(x)v'(x)] / [v(x)]^2. It might look a bit intimidating at first, but once you break it down, it's quite manageable. In our case, u(x) = x - 1 and v(x) = x + 1. So, the first step is to find the derivatives of u(x) and v(x) separately. The derivative of u(x) = x - 1 is simply u'(x) = 1, and the derivative of v(x) = x + 1 is v'(x) = 1. Now we have all the pieces we need to apply the quotient rule. Plugging everything into the formula, we get: f'(x) = [(1)(x + 1) - (x - 1)(1)] / (x + 1)^2. Simplifying the numerator, we have f'(x) = (x + 1 - x + 1) / (x + 1)^2, which further simplifies to f'(x) = 2 / (x + 1)^2. Awesome! We've found the derivative of our function. But remember, we're not done yet. We want to find the slope of the tangent line at x = 0. So, the final step is to substitute x = 0 into our derivative f'(x). This gives us f'(0) = 2 / (0 + 1)^2 = 2 / 1 = 2. And there you have it! The slope of the tangent line to the function f(x) = (x - 1) / (x + 1) at x = 0 is 2. So, the correct answer is B. 2.

Finding the Derivative of f(x) = 1/x

Next up, let's find the derivative of f(x) = 1/x. This one is a bit more straightforward but still a fundamental concept in calculus. Now, how do we tackle this? Well, there are a couple of ways we could approach it. One option is to use the power rule, but first, we need to rewrite f(x) in a more suitable form. Remember that 1/x can be written as x^(-1). This is a crucial step because it allows us to apply the power rule directly. The power rule states that if we have a function f(x) = x^n, where n is any real number, then its derivative f'(x) is given by f'(x) = nx^(n-1). In simpler terms, you multiply by the exponent and then subtract 1 from the exponent. Easy peasy! Applying the power rule to f(x) = x^(-1), we get f'(x) = (-1)x^(-1-1) = -x^(-2). But we're not quite finished yet. We usually want to express our derivatives with positive exponents. So, we can rewrite -x^(-2) as -1/x^2. And there we have it! The derivative of f(x) = 1/x is f'(x) = -1/x^2. Another way to find this derivative is using the quotient rule, even though it might seem a bit like overkill for such a simple function. But it's a good exercise to see how the quotient rule applies in different situations. If we treat f(x) = 1/x as u(x) / v(x), where u(x) = 1 and v(x) = x, then we can apply the quotient rule just like we did in the previous problem. The derivative of u(x) = 1 is u'(x) = 0, and the derivative of v(x) = x is v'(x) = 1. Plugging these into the quotient rule formula, we get: f'(x) = [(0)(x) - (1)(1)] / x^2. Simplifying, we have f'(x) = -1 / x^2, which is exactly the same answer we got using the power rule. So, whether you use the power rule or the quotient rule, the derivative of f(x) = 1/x is indeed f'(x) = -1/x^2. The correct answer here is B. -1/x^2.

Key Concepts and Takeaways

Alright, let's recap what we've learned today. We've covered two important calculus concepts: finding the slope of a tangent line and calculating derivatives. We saw that the slope of the tangent line to a function at a specific point is given by the derivative of the function evaluated at that point. This is a fundamental connection between derivatives and the geometry of curves. To find the derivative of f(x) = (x - 1) / (x + 1), we used the quotient rule, which is essential for differentiating functions that are the ratio of two other functions. The quotient rule formula might look intimidating at first, but with practice, it becomes second nature. Remember the formula: f'(x) = [u'(x)v(x) - u(x)v'(x)] / [v(x)]^2. Breaking down the problem into smaller steps – identifying u(x) and v(x), finding their derivatives, and then plugging everything into the formula – makes it much more manageable. For finding the derivative of f(x) = 1/x, we explored two methods: the power rule and the quotient rule. The power rule is a super handy shortcut when you can rewrite your function in the form x^n. By rewriting 1/x as x^(-1), we could directly apply the power rule and quickly find the derivative. The quotient rule also works, although it might be a bit more steps for this particular function. The key takeaway here is that there are often multiple ways to solve a calculus problem, and choosing the most efficient method can save you time and effort. Both the power rule and the quotient rule are essential tools in your calculus toolbox, so make sure you're comfortable using them. Practice makes perfect, so try applying these rules to other functions and see how you go!

Wrapping Up

So, guys, we've successfully navigated through finding the slope of a tangent line and calculating derivatives today. We've seen how the quotient rule and the power rule can be used to differentiate various functions. Remember, calculus might seem daunting at first, but with a solid understanding of the fundamental concepts and a bit of practice, you can tackle even the trickiest problems. Keep practicing, keep exploring, and most importantly, keep having fun with math! If you have any questions or want to explore other calculus topics, feel free to ask. Until next time, happy calculating!