Hey guys! Today, let's dive into a classic calculus problem: finding the derivative of the square root function, f(x) = √x, where x > 0. This is a fundamental concept in calculus, and understanding it will help you tackle more complex problems down the road. So, grab your pencils, and let's get started!
Understanding the Derivative
Before we jump into the nitty-gritty, let's quickly recap what a derivative actually is. In simple terms, the derivative of a function at a point represents the instantaneous rate of change of the function at that point. Think of it as the slope of the line tangent to the function's graph at that specific location. This slope tells us how much the function's output is changing in response to a tiny change in its input. Derivatives are the cornerstone of differential calculus, used extensively in physics, engineering, economics, and many other fields to model and optimize dynamic systems.
The Limit Definition of the Derivative
The most fundamental way to define the derivative is using the limit definition. The limit definition of the derivative provides us with a rigorous way to calculate the derivative. For a function f(x), its derivative, denoted as f'(x), is defined as:
f'(x) = lim (h->0) [f(x + h) - f(x)] / h
This formula essentially calculates the slope of a secant line between two points on the function's graph, x and x + h, and then takes the limit as h approaches zero. As h gets infinitesimally small, the secant line becomes the tangent line, and its slope becomes the derivative at the point x. This limit definition is incredibly powerful because it allows us to find derivatives from first principles, without relying on pre-established rules.
Why x > 0?
You might be wondering why we're restricting x to be greater than 0. Well, the square root function, √x, is only defined for non-negative values of x. We can't take the square root of a negative number and get a real result. Furthermore, the derivative of √x is not defined at x = 0 because the function has a vertical tangent at that point. Think about the graph of √x – it starts at the origin and curves upwards, becoming steeper and steeper as it approaches zero. This infinite steepness means the derivative doesn't exist at x = 0. Therefore, we focus on the domain where x > 0 to ensure we're working with real numbers and a well-defined derivative.
Applying the Limit Definition to f(x) = √x
Alright, now that we've got the background sorted, let's roll up our sleeves and apply the limit definition to find the derivative of f(x) = √x. This involves a bit of algebraic manipulation, but don't worry, we'll break it down step by step.
Step 1: Setting up the Limit
First, we need to plug our function, f(x) = √x, into the limit definition of the derivative:
f'(x) = lim (h->0) [√(x + h) - √x] / h
This expression looks a bit intimidating, but we're going to simplify it using a clever trick. Notice that we have a difference of square roots in the numerator. This is a classic situation where we can use the conjugate to rationalize the numerator.
Step 2: Rationalizing the Numerator
The conjugate of √(x + h) - √x is √(x + h) + √x. We multiply the numerator and denominator of our expression by this conjugate:
f'(x) = lim (h->0) {[√(x + h) - √x] / h} * {[√(x + h) + √x] / [√(x + h) + √x]}
Multiplying by the conjugate might seem like we're making things more complicated, but it's actually going to help us eliminate the square roots in the numerator. When we multiply the numerators, we get a difference of squares:
[√(x + h) - √x] * [√(x + h) + √x] = (x + h) - x
This simplifies beautifully to just h! Our expression now looks like this:
f'(x) = lim (h->0) h / {h[√(x + h) + √x]}
Step 3: Simplifying the Expression
Now we can cancel the h in the numerator and denominator, which is a crucial step because it removes the h from the denominator, preventing us from dividing by zero when we take the limit:
f'(x) = lim (h->0) 1 / [√(x + h) + √x]
This is much simpler! We've successfully eliminated the troublesome h in the denominator.
Step 4: Evaluating the Limit
Finally, we can evaluate the limit by letting h approach 0:
f'(x) = 1 / [√(x + 0) + √x] = 1 / (√x + √x) = 1 / (2√x)
And there we have it! The derivative of f(x) = √x is f'(x) = 1 / (2√x).
The Derivative: f'(x) = 1 / (2√x)
So, after all that work, we've found that the derivative of f(x) = √x is f'(x) = 1 / (2√x). This formula tells us the slope of the tangent line to the graph of √x at any point x in its domain (x > 0).
Interpreting the Result
Let's think about what this derivative means. The derivative f'(x) = 1 / (2√x) is always positive for x > 0, which means the function √x is always increasing. This makes sense if you picture the graph of √x – it's a curve that starts at the origin and rises gradually as x increases.
Also, notice that as x gets larger, the derivative f'(x) gets smaller. This tells us that the rate of change of √x decreases as x increases. In other words, the graph of √x becomes less steep as we move to the right. At small values of x, √x increases rapidly, but as x grows, the increase becomes more gradual.
Domain of the Derivative
It's important to note that the derivative f'(x) = 1 / (2√x) is also only defined for x > 0. Just like the original function, we can't take the square root of a negative number. Additionally, the derivative is undefined at x = 0 because we would be dividing by zero. This reinforces our earlier point that the derivative of √x doesn't exist at the origin due to the vertical tangent.
Why This Matters: Applications of the Derivative
Now, you might be thinking,