Finding F'(0) For F(x) = E^(1 - 2x) * Cos 2x A Step-by-Step Solution

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Hey guys! Today, we're diving into a calculus problem where we need to find the derivative of a function and then evaluate it at a specific point. Specifically, we're tasked with finding f'(0) given that f(x) = e^(1 - 2x) * cos 2x. This problem involves a mix of exponential and trigonometric functions, so we'll need to use the product rule and chain rule. Let's break it down step by step to make sure we nail it.

Understanding the Problem

Before we jump into the solution, let's make sure we understand what the problem is asking. We have a function, f(x) = e^(1 - 2x) * cos 2x, and we need to find its derivative, f'(x). Once we have the derivative, we need to evaluate it at x = 0 to find f'(0). This means we'll be substituting 0 for x in our derivative function. This type of problem is a classic example of what you might see in a calculus course, and it’s super important to get comfortable with these steps. So, let’s get started!

Why This Matters

Understanding derivatives is crucial in calculus. The derivative, f'(x), tells us the instantaneous rate of change of the function f(x). In simpler terms, it tells us how much the function's output changes for a tiny change in the input. This concept is used everywhere from physics (calculating velocity and acceleration) to economics (analyzing marginal cost and revenue). Mastering these types of problems builds a solid foundation for more advanced topics in mathematics and its applications.

Step 1 The Product Rule

Our function f(x) = e^(1 - 2x) * cos 2x is a product of two functions: u(x) = e^(1 - 2x) and v(x) = cos 2x. To find the derivative of a product, we use the product rule. The product rule states that if we have a function h(x) = u(x) * v(x), then its derivative h'(x) is given by:

h'(x) = u'(x) * v(x) + u(x) * v'(x)

This is a fundamental rule in calculus, and it's essential for differentiating functions that are products of other functions. Think of it as breaking down a complex problem into smaller, manageable parts. In our case, we need to find the derivatives of u(x) and v(x) separately and then plug them into the product rule formula. This approach makes the overall differentiation process much smoother and less prone to errors.

Step 2 Finding u'(x)

Let's find the derivative of u(x) = e^(1 - 2x). This involves using the chain rule, since we have a function inside another function. The chain rule states that if we have a function y = f(g(x)), then its derivative y' is given by:

y' = f'(g(x)) * g'(x)

In our case, we can think of u(x) as e^w, where w = 1 - 2x. So, we need to find the derivative of e^w with respect to w and then multiply it by the derivative of w with respect to x. The derivative of e^w with respect to w is simply e^w. Now we need to find the derivative of w = 1 - 2x with respect to x. This is straightforward:

w' = -2

Now, applying the chain rule, we get:

u'(x) = e^(1 - 2x) * (-2) = -2e^(1 - 2x)

So, we’ve found the derivative of our first component. This step is crucial because the chain rule is a common pitfall for many students. By breaking it down and understanding each part, we ensure we get the correct derivative. Now, let’s move on to the next part.

Step 3 Finding v'(x)

Next, we need to find the derivative of v(x) = cos 2x. Again, we'll use the chain rule. We can think of v(x) as cos(w), where w = 2x. The derivative of cos(w) with respect to w is -sin(w). The derivative of w = 2x with respect to x is:

w' = 2

Applying the chain rule, we get:

v'(x) = -sin(2x) * 2 = -2sin(2x)

Great! We've found the derivative of the second component. The chain rule is really the star of the show here, and understanding how to apply it correctly is key to solving these types of problems. Now that we have both u'(x) and v'(x), we can put everything together using the product rule.

Step 4 Applying the Product Rule

Now that we have u'(x) = -2e^(1 - 2x) and v'(x) = -2sin(2x), we can apply the product rule:

f'(x) = u'(x) * v(x) + u(x) * v'(x)

Plug in the derivatives and the original functions:

f'(x) = (-2e^(1 - 2x)) * (cos 2x) + (e^(1 - 2x)) * (-2sin 2x)

Now, let's simplify this expression:

f'(x) = -2e^(1 - 2x)cos 2x - 2e^(1 - 2x)sin 2x

We can factor out -2e^(1 - 2x) from both terms:

f'(x) = -2e^(1 - 2x)(cos 2x + sin 2x)

Alright, we've got our derivative! This step is where all our hard work comes together. By carefully applying the product rule and simplifying the expression, we’ve arrived at a manageable form for f'(x). Now, the final step is to evaluate this at x = 0.

Step 5 Evaluating f'(0)

Finally, we need to find f'(0). This means substituting x = 0 into our derivative function:

f'(0) = -2e^(1 - 20)(cos(20) + sin(2*0))

Simplify the expression:

f'(0) = -2e^(1)(cos(0) + sin(0))

We know that cos(0) = 1 and sin(0) = 0, so:

f'(0) = -2e(1 + 0)

f'(0) = -2e

So, the answer is -2e, which corresponds to option A.

Breaking Down the Final Step

Evaluating f'(0) might seem like a simple substitution, but it's a critical step. It’s where we go from having a general expression for the derivative to a specific value. The key here is to remember the values of trigonometric functions at special angles like 0. Cosine of 0 is 1, and sine of 0 is 0. Plugging these values in, we get a straightforward calculation that leads us to our final answer.

Conclusion

We've successfully found f'(0) for the given function. We used the product rule and the chain rule, and we carefully evaluated the result at x = 0. The correct answer is A) -2e. Remember, the key to mastering calculus is practice. Work through similar problems, and you'll become more confident in your skills. You got this!

Final Thoughts and Tips

Calculus problems like this one might seem daunting at first, but they become much easier with practice. Here are a few final tips to keep in mind:

  • Master the Rules: Make sure you have a solid understanding of the product rule, quotient rule, and chain rule. These are the fundamental tools for differentiation.
  • Break It Down: Complex functions can be broken down into simpler components. Identify the inner and outer functions when using the chain rule.
  • Practice Regularly: The more you practice, the more comfortable you’ll become with applying these rules.
  • Check Your Work: Always double-check your derivatives and substitutions to avoid simple errors.

And that's it, guys! We've tackled a challenging calculus problem together. Keep practicing, and you’ll be a calculus pro in no time! Happy calculating!