Calculating F'(π): A Calculus Deep Dive

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Hey everyone, let's dive into a fun calculus problem! We're going to calculate the value of the derivative of a function at a specific point. Specifically, we're dealing with the function f(x) = sin(2x) / √x + 6 and we want to find f'(π). Don't worry, it sounds more intimidating than it is. We'll break it down step-by-step, making sure everyone understands the process. This isn't just about getting the answer; it's about understanding the how and why behind it. Ready to get started, guys?

Understanding the Problem and Tools

Alright, so what does this even mean? We're given a function, f(x). We need to find its derivative, which we denote as f'(x). The derivative represents the instantaneous rate of change of the function at any given point. Think of it as the slope of the tangent line to the curve of the function at that point. Once we have the derivative, we'll plug in π (pi, or approximately 3.14159) for x to find the specific value of the derivative at that point. This involves a couple of key calculus concepts:

  • Differentiation: This is the process of finding the derivative of a function. We'll need to use the rules of differentiation. In our case, we'll need to use the quotient rule and the chain rule. The quotient rule helps us differentiate a function that's a fraction (one function divided by another). The chain rule is necessary because we have a composite function (a function within another function).
  • Trigonometry: Since our function involves sin(2x), a basic understanding of sine functions is useful. Although, the differentiation rules will handle the bulk of the work. You'll want to know the derivative of sin(x), which is cos(x).
  • Algebraic Manipulation: We'll need to be comfortable simplifying expressions after we've taken the derivative. This includes things like combining terms and evaluating the function at x = π.

Before we jump in, let's recall some essential derivative rules:

  • Power Rule: The derivative of xⁿ is nxⁿ⁻¹*. For example, the derivative of x² is 2x.
  • Derivative of sin(x): The derivative of sin(x) is cos(x).
  • Chain Rule: If we have a composite function, like f(g(x)), the derivative is f'(g(x)) * g'(x). This is super important for us.
  • Quotient Rule: If we have a function in the form of u(x) / v(x), the derivative is (u'(x)v(x) - u(x)v'(x)) / [v(x)]².

With these tools in our toolbox, we are ready to proceed. Take a deep breath and let's get to work. This is going to be fun, I promise!

Step-by-Step Differentiation of f(x)

Okay, let's start by differentiating the function f(x) = sin(2x) / √x + 6. We can rewrite this as f(x) = sin(2x) / (x^(1/2) + 6) to make the differentiation process a bit clearer. This function has a quotient structure, so we'll need to employ the quotient rule. Let's define:

  • u(x) = sin(2x) (the numerator)
  • v(x) = x^(1/2) + 6 (the denominator)

Now, let's find the derivatives of u(x) and v(x).

  1. Differentiating u(x) = sin(2x): This is where the chain rule comes in! The derivative of sin(u) is cos(u) * u'. In our case, u = 2x, so u' = 2. Therefore, u'(x) = cos(2x) * 2 = 2cos(2x).
  2. Differentiating v(x) = x^(1/2) + 6: Using the power rule, the derivative of x^(1/2) is (1/2)x^(-1/2). The derivative of a constant (like 6) is 0. Therefore, v'(x) = (1/2)x^(-1/2) or, equivalently, v'(x) = 1 / (2√x).

Now we have all the pieces needed for the quotient rule. Let's put it all together.

Applying the Quotient Rule: The quotient rule states that if f(x) = u(x) / v(x), then f'(x) = (u'(x)v(x) - u(x)v'(x)) / [v(x)]². Let's substitute what we know:

  • f'(x) = (2cos(2x) * (x^(1/2) + 6) - sin(2x) * (1 / (2√x))) / (x^(1/2) + 6)²

Simplify the result: This is our derivative! Now, let's simplify it a bit. We'll multiply through and try to make it look a little cleaner:

  • f'(x) = (2cos(2x)√x + 12cos(2x) - sin(2x) / (2√x)) / (x + 12√x + 36) (I expanded the denominator too).

That's the derivative, guys! The next step is to plug in x = π.

Evaluating f'(π)

Now, the fun part: substituting π for x in our derivative f'(x). Let's substitute, being mindful of the trigonometric function and the square roots:

  • f'(π) = (2cos(2π)√π + 12cos(2π) - sin(2π) / (2√π)) / (π + 12√π + 36).

Let's simplify this step by step.

  1. Evaluating Trigonometric Functions: We know that cos(2π) = 1 and sin(2π) = 0. This simplifies our expression considerably.
  2. Substituting the values: Substituting these trigonometric values into the equation, we get: f'(π) = (2 * 1 * √π + 12 * 1 - 0 / (2√π)) / (π + 12√π + 36).
  3. Simplifying Further: This simplifies down to: f'(π) = (2√π + 12) / (π + 12√π + 36).

So, the value of f'(π) is (2√π + 12) / (π + 12√π + 36). We can approximate this. Since π ≈ 3.14159, then √π ≈ 1.772. Plugging this into the formula, we get: f'(π) ≈ (21.772 + 12) / (3.14159 + 121.772 + 36) ≈ 15.544/55.415 ≈ 0.2805.

Conclusion and Key Takeaways

We did it! We found that f'(π) = (2√π + 12) / (π + 12√π + 36). We also approximated this value as approximately 0.2805. We used a lot of different rules of differentiation and made sure to not make any mistakes. This calculation involved the quotient rule and the chain rule, along with a good understanding of trigonometric functions and basic algebraic manipulations. Remember, calculus is all about understanding how things change, and the derivative is our tool to measure that change. I hope this explanation helped you to better understand derivatives, the application of differentiation rules, and how to evaluate a derivative at a specific point. Keep practicing, and these concepts will become second nature! You guys are awesome.

Key Takeaways:

  • The Quotient Rule: Remember the formula and when to apply it. It's essential for functions in the form of a fraction.
  • The Chain Rule: This is your go-to rule for composite functions. Master it!
  • Trigonometric Derivatives: Know the derivatives of sin(x) and cos(x).
  • Step-by-Step Approach: Break down the problem into smaller, manageable steps. It makes the process a lot less daunting.
  • Practice: The more you practice, the better you'll get. Try similar problems to solidify your understanding.

Keep up the great work, and happy calculating, everyone! Let me know if you have any questions in the comments below! Good job, guys!