Calculating Derivatives: A Step-by-Step Guide
Hey guys! Let's dive into the world of derivatives, specifically calculating them using the definition. It might sound a bit intimidating at first, but trust me, we'll break it down into easy-to-understand steps. We're going to tackle four different functions and find their derivatives at specific points. This process will give you a solid grasp of what a derivative actually is – the instantaneous rate of change of a function. So, grab your pencils and calculators (or your favorite math software), and let's get started! We'll go through each example step-by-step, making sure you understand every move.
2.1. Derivative of f(x) = x² at x₀ = 2
Alright, let's kick things off with f(x) = x² and find its derivative at the point x₀ = 2. To do this using the definition, we'll use the following formula:
f'(x₀) = lim (h → 0) [f(x₀ + h) - f(x₀)] / h
This formula is the heart of the definition of a derivative. It represents the limit of the slope of the secant line as the change in x (represented by h) approaches zero. In simpler terms, we're finding the slope of the tangent line at the point x₀. It's like zooming in on the curve and seeing how it behaves at that single point.
First, let's calculate f(x₀). Since x₀ = 2, we have f(2) = 2² = 4. Now, let's find f(x₀ + h). This means replacing x in our function with (x₀ + h), which is (2 + h). So, f(2 + h) = (2 + h)² = 4 + 4h + h². Remembering the binomial expansion? It's a good friend in these calculations.
Now, we plug these values into our derivative formula:
f'(2) = lim (h → 0) [(4 + 4h + h²) - 4] / h
Simplify the numerator: (4 + 4h + h²) - 4 = 4h + h². Our expression now looks like this:
f'(2) = lim (h → 0) [4h + h²] / h
We can factor out an h from the numerator:
f'(2) = lim (h → 0) h(4 + h) / h
Now, we can cancel the h in the numerator and denominator (as long as h is not equal to zero – which it isn't, because we are looking at the limit as h approaches zero, not equal to zero). This leaves us with:
f'(2) = lim (h → 0) (4 + h)
Finally, let h approach zero. This gives us f'(2) = 4 + 0 = 4. So, the derivative of f(x) = x² at x₀ = 2 is 4. This means the slope of the tangent line to the curve f(x) = x² at the point (2, 4) is 4. This is a super important concept! It tells us how the function is changing at that specific point. Pretty cool, right?
2.2. Derivative of f(x) = x³ at x₀ = -2
Let's move on to our second example: finding the derivative of f(x) = x³ at x₀ = -2. We'll use the same definition formula as before. This time, it's slightly different, but the process remains the same.
First, let's find f(x₀). Since x₀ = -2, we have f(-2) = (-2)³ = -8. Next, we need to find f(x₀ + h). This means f(-2 + h) = (-2 + h)³. To expand this, we can use the binomial theorem or just multiply it out step by step: (-2 + h) * (-2 + h) * (-2 + h) = (4 - 4h + h²) * (-2 + h) = -8 + 12h - 6h² + h³.
Now, plug these values into the derivative formula:
f'(-2) = lim (h → 0) [(-8 + 12h - 6h² + h³) - (-8)] / h
Simplify the numerator: (-8 + 12h - 6h² + h³) - (-8) = 12h - 6h² + h³. Our expression now looks like this:
f'(-2) = lim (h → 0) [12h - 6h² + h³] / h
Factor out an h from the numerator:
f'(-2) = lim (h → 0) h(12 - 6h + h²) / h
Cancel the h in the numerator and denominator:
f'(-2) = lim (h → 0) (12 - 6h + h²)
Let h approach zero. This gives us f'(-2) = 12 - 6(0) + 0² = 12. So, the derivative of f(x) = x³ at x₀ = -2 is 12. This means the slope of the tangent line to the curve f(x) = x³ at the point (-2, -8) is 12. Notice how we are always looking at the instantaneous rate of change. That's the essence of the derivative! Understanding this concept is crucial for understanding calculus. We're essentially finding how the function is changing at a specific point on the curve. It's like having a magnifying glass and zooming in on the graph.
2.3. Derivative of f(x) = eˣ at x₀ = 0
Now, let's calculate the derivative of the exponential function f(x) = eˣ at x₀ = 0. This one involves the exponential function e, which is a fundamental constant in mathematics. The process, however, is the same!
First, find f(x₀). Since x₀ = 0, we have f(0) = e⁰ = 1. Then, we find f(x₀ + h), which is f(0 + h) = e^(0+h) = e^h.
Substitute these values into the derivative formula:
f'(0) = lim (h → 0) [e^h - 1] / h
This limit is a bit tricky. It requires a special limit result that we'll simply state here:
lim (h → 0) [e^h - 1] / h = 1
Therefore, f'(0) = 1. This means the derivative of f(x) = eˣ at x₀ = 0 is 1. The tangent line to the curve of eˣ at the point (0, 1) has a slope of 1. This is a special property of the exponential function: its derivative at x = 0 is equal to 1. Isn't that neat? The derivative of eˣ is actually eˣ itself!
2.4. Derivative of f(x) = sin(x) at x₀ = π
Finally, let's tackle the derivative of the sine function, f(x) = sin(x), at x₀ = π. This one involves trigonometric functions, but don't worry, the process is the same.
First, find f(x₀). Since x₀ = π, we have f(π) = sin(π) = 0. Then, find f(x₀ + h), which is f(π + h) = sin(π + h).
Now, we need to use the trigonometric identity sin(π + h) = -sin(h). So, f(π + h) = -sin(h).
Substitute these values into the derivative formula:
f'(π) = lim (h → 0) [-sin(h) - 0] / h
Simplify the expression:
f'(π) = lim (h → 0) [-sin(h) / h]
We use another special limit result here:
lim (h → 0) sin(h) / h = 1
Therefore, f'(π) = -1. So, the derivative of f(x) = sin(x) at x₀ = π is -1. The tangent line to the curve of sin(x) at the point (π, 0) has a slope of -1. This makes sense because the sine function is decreasing at x = π. This means that at the point (π,0), the function is going down. This confirms our answer of -1. This is another example of how derivatives help us understand the behavior of functions!
That's it, guys! We've successfully calculated the derivatives of four different functions using the definition. Remember that the definition of the derivative is fundamental to understanding calculus. Keep practicing, and you'll get the hang of it in no time. Mastering the definition opens the door to understanding all the derivative rules.