Calculus Breakdown: Derivatives And Domains Of Logarithmic Functions

Hey math enthusiasts! Let's dive into a cool calculus problem today. We're going to break down a function, find its derivative, and figure out where it "lives." Specifically, we're looking at the function f(x) = x² * ln(10 - 4x²). So, grab your pencils (or your favorite digital stylus), and let's get started!

Unveiling the Derivative: Finding f'(x)

Alright, guys, the first part of our mission is to find the derivative of f(x). Remember, the derivative tells us the rate of change of a function. In simpler terms, it gives us the slope of the tangent line at any point on the curve. This problem involves a combination of the product rule and the chain rule, so let's recall what these rules are.

The product rule states that if we have a function that's the product of two other functions, say u(x) and v(x), then the derivative of the whole thing is: (uv)' = u'v + uv'. Basically, it means the derivative of the first function times the second function, plus the first function times the derivative of the second function.

The chain rule comes into play when we have a function inside another function (a composite function). If we have a function g(x) inside h(x), so we have h(g(x)), the derivative is: (h(g(x)))' = h'(g(x)) * g'(x). We find the derivative of the outer function, evaluate it at the inner function, and then multiply by the derivative of the inner function.

Now, let's look at our specific function, f(x) = x² * ln(10 - 4x²). We can see that it's a product of two functions: u(x) = x² and v(x) = ln(10 - 4x²). So, we'll need to use the product rule first.

Let's find the derivatives of u(x) and v(x).

• u'(x) is straightforward; the derivative of x² is 2x.
• For v'(x), we'll need to use the chain rule. v(x) = ln(10 - 4x²). The derivative of ln(x) is 1/x. So, the derivative of the outside function (natural log) is 1/(10 - 4x²). Then we multiply by the derivative of the inside function (10 - 4x²), which is -8x. Therefore, v'(x) = (-8x) / (10 - 4x²).

Now, let's put it all together using the product rule: f'(x) = u'v + uv'. This becomes f'(x) = (2x) * ln(10 - 4x²) + x² * ((-8x) / (10 - 4x²)). Simplifying this, we get f'(x) = 2x ln(10 - 4x²) - (8x³ / (10 - 4x²)). There you have it, folks! We've successfully found the derivative of f(x).

To make this a little prettier, we could even combine those two terms. You would need to put them over a common denominator, but I'll leave the algebra to you. That's our final answer for f'(x).

Unraveling the Domain: Where f(x) Exists

Next up, we need to find the domain of our function f(x) = x² * ln(10 - 4x²). The domain is simply the set of all x-values for which the function is defined. Because we have a natural logarithm (ln) in our function, we need to consider some constraints.

Remember, the natural logarithm, ln(x), is only defined for positive values of x. We cannot take the natural log of zero or a negative number. This is crucial for determining our domain. So, for our function, the argument of the natural logarithm, which is (10 - 4x²), must be greater than zero. Mathematically, we need to solve the inequality: 10 - 4x² > 0.

Let's go through the steps to solve this inequality:

1. Isolate the x² term: Subtract 10 from both sides: -4x² > -10.
2. Divide both sides by -4: Remember, when you divide or multiply both sides of an inequality by a negative number, you need to flip the inequality sign. So, we get x² < 10/4, which simplifies to x² < 5/2.
3. Take the square root of both sides: When taking the square root, we need to consider both positive and negative roots. So, we have -√(5/2) < x < √(5/2).

This means that x must be greater than the negative square root of 5/2 and less than the positive square root of 5/2. In other words, x can be any value between -√(5/2) and √(5/2). These are approximately -1.58 and 1.58.

Using interval notation, the domain of f(x) is (-√(5/2), √(5/2)). We use parentheses because the values -√(5/2) and √(5/2) are not included in the domain (because the natural log isn't defined at zero). Any value of x inside of this interval will give us a valid output for our function, which is the definition of the domain.

Wrapping It Up: Key Takeaways

Alright, guys, let's recap what we've learned:

• We found the derivative f'(x) = 2x ln(10 - 4x²) - (8x³ / (10 - 4x²)), using both the product rule and the chain rule.
• We determined the domain of f(x) to be (-√(5/2), √(5/2)), which is the set of all x-values for which the function is defined.

These are pretty fundamental concepts in calculus, so make sure you understand them well. Practice makes perfect, so keep working through problems like this, and you'll become a calculus whiz in no time!

This type of problem emphasizes the importance of understanding derivative rules and how to apply them. Additionally, it highlights the importance of understanding the limitations of the functions we work with, in this case, the domain of a logarithmic function.

Keep practicing and you'll conquer calculus, one step at a time! And don't be afraid to ask for help; that's what we're all here for.