Derivative Of F(x) = X² Ln(5x² + 2x⁻³): A Step-by-Step Guide

Oct 20, 2025 by ADMIN 61 views

Hey guys! Today, we're diving into a fun calculus problem: finding the derivative of the function f(x) = x² ln(5x² + 2x⁻³). This might look a little intimidating at first, but don't worry! We'll break it down step-by-step, using the product rule and the chain rule. So, grab your pencils, and let's get started!

Understanding the Problem

Before we jump into the solution, let's quickly understand what we're dealing with. We have a function, f(x), which is a product of two terms: x² and ln(5x² + 2x⁻³). This means we'll definitely need the product rule. Also, notice that the second term involves a natural logarithm with a more complex expression inside, so the chain rule will be our friend here too.

Why is this important? Derivatives are fundamental in calculus, representing the instantaneous rate of change of a function. They help us understand how a function is behaving, whether it's increasing, decreasing, or staying constant. Finding derivatives has applications in various fields, from physics and engineering to economics and computer science. Think about it – optimizing processes, modeling growth, or even predicting stock prices often involve derivatives! So, mastering this skill is super valuable.

Let's break down the key concepts we'll be using:

Product Rule: This rule helps us differentiate a product of two functions. If we have two functions, u(x) and v(x), then the derivative of their product is: (uv)' = u'v + uv'
Chain Rule: This rule is used when we have a composite function (a function within a function). If we have f(g(x)), then its derivative is: f'(g(x)) * g'(x)

With these rules in our toolbox, we're ready to tackle the problem!

Applying the Product Rule

Our function, as we've seen, is f(x) = x² ln(5x² + 2x⁻³). Let's identify our u(x) and v(x):

u(x) = x²
v(x) = ln(5x² + 2x⁻³)

Now, we need to find the derivatives of u(x) and v(x). The derivative of u(x) is straightforward:

u'(x) = 2x

Finding the derivative of v(x) is where the chain rule comes in. We have a natural logarithm function with an inner function (5x² + 2x⁻³). So, let's break it down further:

Let g(x) = 5x² + 2x⁻³

Then v(x) = ln(g(x))

Now we can apply the chain rule. The derivative of ln(x) is 1/x, so the derivative of ln(g(x)) is (1/g(x)) * g'(x). We need to find g'(x):

g'(x) = d/dx (5x² + 2x⁻³)

Using the power rule, we get:

g'(x) = 10x - 6x⁻⁴

Now we can find v'(x):

v'(x) = (1/g(x)) * g'(x) v'(x) = (1 / (5x² + 2x⁻³)) * (10x - 6x⁻⁴)

Okay, we've got u(x), u'(x), v(x), and v'(x). It's time to plug everything into the product rule formula: (uv)' = u'v + uv'

f'(x) = (2x) * ln(5x² + 2x⁻³) + (x²) * ((1 / (5x² + 2x⁻³)) * (10x - 6x⁻⁴))

Simplifying the Derivative

Alright, we've found the derivative, but it looks a bit messy. Let's simplify it to make it more manageable. Our current expression is:

f'(x) = 2x ln(5x² + 2x⁻³) + x² * (10x - 6x⁻⁴) / (5x² + 2x⁻³)

We can rewrite the second term to get a common denominator:

f'(x) = 2x ln(5x² + 2x⁻³) + (10x³ - 6x⁻²) / (5x² + 2x⁻³)

To further simplify, we can multiply the numerator and denominator of the second term by x³ to get rid of the negative exponent:

f'(x) = 2x ln(5x² + 2x⁻³) + (10x⁶ - 6) / (x³(5x² + 2x⁻³))

f'(x) = 2x ln(5x² + 2x⁻³) + (10x⁶ - 6) / (5x⁵ + 2)

This is a much cleaner expression for the derivative. We've successfully applied the product rule and the chain rule and simplified the result.

Why is simplification important? A simplified derivative is not only easier to read but also much easier to work with in further calculations. For example, if you needed to find critical points (where the derivative is zero or undefined) or analyze the concavity of the function, a simplified expression will make your life a whole lot easier.

Common Mistakes to Avoid

When working with derivatives, especially with the product and chain rules, there are a few common mistakes people often make. Let's look at some of them so you can avoid falling into these traps:

Forgetting the Chain Rule: A big mistake is forgetting to apply the chain rule when differentiating a composite function. Remember, if you have a function inside another function, you need to multiply by the derivative of the inner function.
Incorrectly Applying the Product Rule: Make sure you're using the correct formula: (uv)' = u'v + uv'. It's easy to mix up the terms or forget one of them.
Algebra Errors in Simplification: Simplifying the derivative can sometimes involve complex algebra. Be careful with your calculations, especially when dealing with fractions and exponents.
Ignoring Negative Exponents: When simplifying, remember that x⁻ⁿ = 1/xⁿ. Failing to handle negative exponents correctly can lead to errors.

Tips to avoid mistakes:

Write down each step: Don't try to do everything in your head. Writing down each step of the process helps you keep track of what you're doing and reduces the chance of making a mistake.
Double-check your work: After you've found the derivative, take a moment to go back and check each step. Look for any potential errors in your calculations or application of the rules.
Practice, practice, practice: The more you practice, the more comfortable you'll become with these concepts and the less likely you are to make mistakes.

Conclusion

So, guys, we've successfully found the derivative of f(x) = x² ln(5x² + 2x⁻³) using the product rule and the chain rule. We've also simplified the result and discussed some common mistakes to avoid. Remember, practice makes perfect, so keep working on these types of problems, and you'll become a derivative master in no time!

Key takeaways:

The product rule is essential for differentiating products of functions.
The chain rule is crucial for composite functions.
Simplifying the derivative is important for easier further calculations.
Avoid common mistakes by writing down each step and double-checking your work.

I hope this step-by-step guide was helpful! If you have any questions, feel free to ask. Now, go out there and conquer those derivatives!