Finding The Derivative Of F(x) = (2x⁴ - 3x + 5)(x² - √x + 2/x)

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Hey everyone! Today, we're diving into a classic calculus problem: finding the derivative of a function that's a product of two expressions. Specifically, we want to find f'(x) if f(x) = (2x⁴ - 3x + 5)(x² - √x + 2/x). This looks a bit intimidating at first, but don't worry, we'll break it down step by step using the product rule and some basic differentiation techniques.

Understanding the Product Rule

The product rule is our best friend when we need to differentiate a function that's formed by multiplying two other functions together. It states that if we have a function f(x) = u(x)v(x), where u(x) and v(x) are differentiable functions, then the derivative f'(x) is given by:

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

In simpler terms, the derivative of the product is the derivative of the first function times the second function, plus the first function times the derivative of the second function. This rule is essential for tackling problems like the one we have here. Think of it as a recipe: we need to identify our 'ingredients' (u(x) and v(x)), find their derivatives (u'(x) and v'(x)), and then plug everything into the formula. Now, let's apply this to our specific problem.

Applying the Product Rule to Our Function

In our case, we can identify our two functions as:

  • u(x) = 2x⁴ - 3x + 5
  • v(x) = x² - √x + 2/x

So, our original function f(x) is simply the product of these two: f(x) = u(x)v(x). Now, we need to find the derivatives of u(x) and v(x). This is where our knowledge of basic differentiation rules comes in handy. Remember the power rule, which states that the derivative of xⁿ is nxⁿ⁻¹? We'll be using that a lot.

Let's start with u(x). We have:

u(x) = 2x⁴ - 3x + 5

To find u'(x), we differentiate each term separately:

  • The derivative of 2x⁴ is 8x³ (using the power rule: 4 * 2 * x^(4-1))
  • The derivative of -3x is -3 (the derivative of x is 1)
  • The derivative of 5 is 0 (the derivative of a constant is always zero)

So, u'(x) = 8x³ - 3. Awesome! We've got one derivative down. Now, let's tackle v(x). This one's a little trickier because of the square root and the term with x in the denominator. But we can handle it by rewriting the terms with fractional and negative exponents. Remember, √x is the same as x^(1/2), and 2/x is the same as 2x⁻¹. So, we can rewrite v(x) as:

v(x) = x² - x^(1/2) + 2x⁻¹

Now we can differentiate each term:

  • The derivative of x² is 2x
  • The derivative of -x^(1/2) is -(1/2)x^(-1/2) (using the power rule: (1/2) * -1 * x^(1/2 - 1))
  • The derivative of 2x⁻¹ is -2x⁻² (using the power rule: -1 * 2 * x^(-1 - 1))

So, v'(x) = 2x - (1/2)x^(-1/2) - 2x⁻². We can also rewrite this using radicals and fractions to make it look a bit cleaner:

v'(x) = 2x - 1/(2√x) - 2/x²

Fantastic! We've found both u'(x) and v'(x). Now we have all the pieces we need to plug into the product rule formula.

Plugging into the Product Rule Formula

Remember the product rule formula: f'(x) = u'(x)v(x) + u(x)v'(x). Let's substitute our expressions for u'(x), v(x), u(x), and v'(x):

f'(x) = (8x³ - 3)(x² - √x + 2/x) + (2x⁴ - 3x + 5)(2x - 1/(2√x) - 2/x²)

Okay, this looks like a mouthful, but we're almost there! This is our derivative, but it's not in the simplest form. To make it look nicer, we need to expand the products and combine like terms. This is where things can get a bit tedious, but it's essential to get to the final simplified answer.

Expanding and Simplifying

Let's start by expanding the first product: (8x³ - 3)(x² - √x + 2/x). We need to multiply each term in the first expression by each term in the second expression:

  • 8x³ * x² = 8x⁵
  • 8x³ * -√x = -8x³x^(1/2) = -8x^(7/2)
  • 8x³ * 2/x = 16x²
  • -3 * x² = -3x²
  • -3 * -√x = 3√x = 3x^(1/2)
  • -3 * 2/x = -6/x = -6x⁻¹

So, the first product expands to: 8x⁵ - 8x^(7/2) + 16x² - 3x² + 3x^(1/2) - 6x⁻¹. Now let's simplify this by combining like terms:

8x⁵ - 8x^(7/2) + 13x² + 3x^(1/2) - 6x⁻¹

Next, we need to expand the second product: (2x⁴ - 3x + 5)(2x - 1/(2√x) - 2/x²). This one's even more involved:

  • 2x⁴ * 2x = 4x⁵
  • 2x⁴ * -1/(2√x) = -x⁴/(√x) = -x^(7/2)
  • 2x⁴ * -2/x² = -4x²
  • -3x * 2x = -6x²
  • -3x * -1/(2√x) = (3/2)x/(√x) = (3/2)x^(1/2)
  • -3x * -2/x² = 6/x = 6x⁻¹
  • 5 * 2x = 10x
  • 5 * -1/(2√x) = -5/(2√x) = -(5/2)x^(-1/2)
  • 5 * -2/x² = -10/x² = -10x⁻²

So, the second product expands to: 4x⁵ - x^(7/2) - 4x² - 6x² + (3/2)x^(1/2) + 6x⁻¹ + 10x - (5/2)x^(-1/2) - 10x⁻². Let's simplify this by combining like terms:

4x⁵ - x^(7/2) - 10x² + (3/2)x^(1/2) + 6x⁻¹ + 10x - (5/2)x^(-1/2) - 10x⁻²

Now, we need to add the two expanded expressions together to get our final derivative:

f'(x) = (8x⁵ - 8x^(7/2) + 13x² + 3x^(1/2) - 6x⁻¹) + (4x⁵ - x^(7/2) - 10x² + (3/2)x^(1/2) + 6x⁻¹ + 10x - (5/2)x^(-1/2) - 10x⁻²)

Finally, let's combine like terms one last time:

  • 8x⁵ + 4x⁵ = 12x⁵
  • -8x^(7/2) - x^(7/2) = -9x^(7/2)
  • 13x² - 10x² = 3x²
  • 3x^(1/2) + (3/2)x^(1/2) = (9/2)x^(1/2)
  • -6x⁻¹ + 6x⁻¹ = 0
  • -(5/2)x^(-1/2) (no like terms)
  • 10x (no like terms)
  • -10x⁻² (no like terms)

So, our final simplified derivative is:

f'(x) = 12x⁵ - 9x^(7/2) + 3x² + (9/2)x^(1/2) + 10x - (5/2)x^(-1/2) - 10x⁻²

We can also rewrite this using radicals and fractions to make it look even cleaner:

f'(x) = 12x⁵ - 9x^(7/2) + 3x² + (9/2)√x + 10x - 5/(2√x) - 10/x²

Conclusion

So there you have it! We've successfully found the derivative of f(x) = (2x⁴ - 3x + 5)(x² - √x + 2/x) using the product rule and some careful simplification. It was a long journey, but we made it! Remember, the key is to break down the problem into smaller steps, apply the rules of differentiation correctly, and take your time with the algebra. You've got this!

Hey guys! Let's break down the process of finding the derivative f'(x) for the function f(x) = (2x⁴ - 3x + 5)(x² - √x + 2/x). This is a classic calculus problem that uses the product rule, so let's take a detailed, step-by-step look at how to solve it. We'll go through each step meticulously, so you can follow along and understand exactly what's happening.

1. Identify the Functions u(x) and v(x)

The first step in using the product rule is to identify the two functions that are being multiplied together. In our case, f(x) is the product of two functions, which we can call u(x) and v(x):

  • u(x) = 2x⁴ - 3x + 5
  • v(x) = x² - √x + 2/x

So, we have f(x) = u(x)v(x). This is the foundation for applying the product rule. Recognizing these components is crucial, as it sets the stage for the next steps. It’s like identifying the ingredients before you start cooking – you need to know what you’re working with!

2. Find the Derivatives u'(x) and v'(x)

Now that we've identified u(x) and v(x), the next step is to find their derivatives, u'(x) and v'(x). This is where our knowledge of basic differentiation rules comes into play. Remember the power rule, which is essential for this process. The power rule states that if f(x) = xⁿ, then f'(x) = nxⁿ⁻¹. We'll also use the fact that the derivative of a constant is zero.

Finding u'(x)

Let's start with u(x) = 2x⁴ - 3x + 5. We'll differentiate each term separately:

  • The derivative of 2x⁴ is 8x³ (using the power rule: 4 * 2 * x^(4-1))
  • The derivative of -3x is -3 (the derivative of x is 1)
  • The derivative of 5 is 0 (the derivative of a constant is always zero)

So, u'(x) = 8x³ - 3. Great! We've found the derivative of the first function. Now, let's move on to v(x).

Finding v'(x)

Finding v'(x) is a bit more challenging because v(x) = x² - √x + 2/x has terms with square roots and a term with x in the denominator. To make differentiation easier, we'll rewrite v(x) using fractional and negative exponents. Remember that √x is the same as x^(1/2), and 2/x is the same as 2x⁻¹. So, we can rewrite v(x) as:

v(x) = x² - x^(1/2) + 2x⁻¹

Now we can differentiate each term:

  • The derivative of x² is 2x
  • The derivative of -x^(1/2) is -(1/2)x^(-1/2) (using the power rule: (1/2) * -1 * x^(1/2 - 1))
  • The derivative of 2x⁻¹ is -2x⁻² (using the power rule: -1 * 2 * x^(-1 - 1))

So, v'(x) = 2x - (1/2)x^(-1/2) - 2x⁻². We can also rewrite this using radicals and fractions to make it look a bit cleaner:

v'(x) = 2x - 1/(2√x) - 2/x²

Excellent! We've found both u'(x) and v'(x). These are the key components we need for the product rule.

3. Apply the Product Rule Formula

Now that we have u(x), v(x), u'(x), and v'(x), we can apply the product rule formula. The product rule states that if f(x) = u(x)v(x), then:

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

Let's substitute our expressions into this formula:

f'(x) = (8x³ - 3)(x² - √x + 2/x) + (2x⁴ - 3x + 5)(2x - 1/(2√x) - 2/x²)

This is our derivative, but it's not in the simplest form. The next step is to expand and simplify this expression.

4. Expand the Products

To simplify f'(x), we need to expand the two products in the expression. This means multiplying each term in the first set of parentheses by each term in the second set of parentheses in both products. This can be a bit tedious, but it's a necessary step to combine like terms and get to the final simplified answer.

Expanding the First Product: (8x³ - 3)(x² - √x + 2/x)

We need to multiply each term in (8x³ - 3) by each term in (x² - √x + 2/x):

  • 8x³ * x² = 8x⁵
  • 8x³ * -√x = -8x³x^(1/2) = -8x^(7/2)
  • 8x³ * 2/x = 16x²
  • -3 * x² = -3x²
  • -3 * -√x = 3√x = 3x^(1/2)
  • -3 * 2/x = -6/x = -6x⁻¹

So, the first product expands to: 8x⁵ - 8x^(7/2) + 16x² - 3x² + 3x^(1/2) - 6x⁻¹. Now let's simplify this by combining like terms:

8x⁵ - 8x^(7/2) + 13x² + 3x^(1/2) - 6x⁻¹

Expanding the Second Product: (2x⁴ - 3x + 5)(2x - 1/(2√x) - 2/x²)

This one is a bit more involved, but we'll follow the same process:

  • 2x⁴ * 2x = 4x⁵
  • 2x⁴ * -1/(2√x) = -x⁴/(√x) = -x^(7/2)
  • 2x⁴ * -2/x² = -4x²
  • -3x * 2x = -6x²
  • -3x * -1/(2√x) = (3/2)x/(√x) = (3/2)x^(1/2)
  • -3x * -2/x² = 6/x = 6x⁻¹
  • 5 * 2x = 10x
  • 5 * -1/(2√x) = -5/(2√x) = -(5/2)x^(-1/2)
  • 5 * -2/x² = -10/x² = -10x⁻²

So, the second product expands to: 4x⁵ - x^(7/2) - 4x² - 6x² + (3/2)x^(1/2) + 6x⁻¹ + 10x - (5/2)x^(-1/2) - 10x⁻². Let's simplify this by combining like terms:

4x⁵ - x^(7/2) - 10x² + (3/2)x^(1/2) + 6x⁻¹ + 10x - (5/2)x^(-1/2) - 10x⁻²

5. Combine Like Terms

Now that we've expanded both products, we need to add them together and combine like terms. This will give us our final simplified derivative.

Let's add the two expanded expressions:

f'(x) = (8x⁵ - 8x^(7/2) + 13x² + 3x^(1/2) - 6x⁻¹) + (4x⁵ - x^(7/2) - 10x² + (3/2)x^(1/2) + 6x⁻¹ + 10x - (5/2)x^(-1/2) - 10x⁻²)

Now, let's combine like terms:

  • 8x⁵ + 4x⁵ = 12x⁵
  • -8x^(7/2) - x^(7/2) = -9x^(7/2)
  • 13x² - 10x² = 3x²
  • 3x^(1/2) + (3/2)x^(1/2) = (9/2)x^(1/2)
  • -6x⁻¹ + 6x⁻¹ = 0
  • -(5/2)x^(-1/2) (no like terms)
  • 10x (no like terms)
  • -10x⁻² (no like terms)

So, our final simplified derivative is:

f'(x) = 12x⁵ - 9x^(7/2) + 3x² + (9/2)x^(1/2) + 10x - (5/2)x^(-1/2) - 10x⁻²

6. Rewrite with Radicals and Fractions (Optional)

We can also rewrite this using radicals and fractions to make it look even cleaner. Remember that x^(1/2) = √x, x^(-1/2) = 1/√x, and x⁻² = 1/x². So, we can rewrite f'(x) as:

f'(x) = 12x⁵ - 9x^(7/2) + 3x² + (9/2)√x + 10x - 5/(2√x) - 10/x²

This is the final, simplified form of the derivative.

Conclusion

We've successfully found f'(x) for the function f(x) = (2x⁴ - 3x + 5)(x² - √x + 2/x). We did it by breaking the problem down into manageable steps: identifying u(x) and v(x), finding their derivatives, applying the product rule, expanding the products, and combining like terms. Remember, practice makes perfect, so keep working on these types of problems, and you'll become a pro at differentiation!

What's up, everyone! When we're tackling derivatives, especially those involving the product rule, it's easy to make a few common mistakes. Today, let's chat about these pitfalls and how to steer clear of them. We'll go over the typical slip-ups people make when using the product rule, and we'll arm you with the knowledge to nail these problems every time. Knowing what to avoid is just as important as knowing the rules themselves!

1. Forgetting the Product Rule Altogether

One of the most common mistakes is simply forgetting that you need to use the product rule in the first place. This usually happens when you see a function that's a product of two expressions, but you try to differentiate it term by term, as if it were a sum or difference. Remember, the product rule is specifically for situations where you have two functions multiplied together.

How to Avoid It: Always double-check the structure of the function before you start differentiating. If you see a product, the product rule should immediately come to mind. Make it a habit to identify the u(x) and v(x) right away. This simple step can save you from a lot of headaches. Think of it like recognizing a traffic sign – you need to know what the sign means before you can react appropriately!

2. Incorrectly Applying the Product Rule Formula

The product rule formula is f'(x) = u'(x)v(x) + u(x)v'(x). It’s crucial to get this formula right. A common mistake is mixing up the terms or forgetting one of them. For instance, some people might write u'(x)v'(x) or u(x) + v(x), which are completely wrong. The order and the plus sign are vital!

How to Avoid It: Write down the formula before you start applying it. This might seem basic, but it's super effective. Having the formula in front of you reduces the chance of making a mistake. Also, practice using the formula regularly. The more you use it, the more it'll stick in your memory. Try saying the formula out loud a few times: "u prime v plus u v prime." It sounds like a little chant, but it helps!

3. Errors in Differentiating u(x) or v(x)

Even if you remember the product rule correctly, you can still make mistakes if you mess up the derivatives of u(x) or v(x). This often happens when these functions involve powers, roots, or fractions. Forgetting the power rule, misapplying the chain rule, or making a simple arithmetic error can lead to an incorrect answer.

How to Avoid It: Take your time when differentiating u(x) and v(x). If they're complex, break them down into smaller steps. For example, if you have a term with a square root, rewrite it using a fractional exponent before differentiating. Double-check each step, and don't rush. It’s better to be accurate than fast. Also, make sure you're solid on your basic differentiation rules before tackling more complex problems.

4. Not Simplifying the Result

Once you've applied the product rule and found f'(x), you're not always done. The problem might ask for the simplified derivative. This means you'll need to expand any products, combine like terms, and possibly rewrite the expression using radicals or fractions. Many people stop after applying the product rule but lose points because they didn't simplify.

How to Avoid It: Always look at the instructions for the problem. If it says "simplify," make sure you do! After applying the product rule, take a moment to see if you can expand and combine terms. It might be helpful to rewrite terms with fractional or negative exponents using radicals or fractions, as this can make simplification easier. Think of it as polishing your work – the final answer should be as clean and clear as possible.

5. Sign Errors

Sign errors are super common in calculus, especially when dealing with multiple terms and operations. A single missed negative sign can throw off the entire solution. This is particularly true when expanding the products after applying the product rule.

How to Avoid It: Be extra careful with your signs. When expanding products, write out each step explicitly, paying close attention to whether terms are positive or negative. Use parentheses to keep track of negative signs, and double-check your work. It might also help to use different colored pens or highlighters to distinguish between positive and negative terms. Small details matter!

Conclusion

Avoiding these common mistakes can significantly improve your accuracy when using the product rule. Remember to recognize when to use the product rule, apply the formula correctly, differentiate u(x) and v(x) carefully, simplify your result, and watch out for those pesky sign errors. Keep practicing, and you'll become a product rule master in no time! You've got this!

Hey everyone! So, we've talked a lot about the product rule – what it is, how to apply it, and what mistakes to avoid. But the best way to truly master a concept in calculus is through practice. That's why I've put together some practice problems for you guys. Working through these will help solidify your understanding of the product rule and give you the confidence to tackle any derivative problem that comes your way. So, grab a pen and paper, and let's get started!

Problem 1

Find f'(x) if f(x) = (x² + 1)(3x - 2)

This is a classic product rule problem. You have two functions, (x² + 1) and (3x - 2), multiplied together. Your first step should be to identify u(x) and v(x), then find their derivatives, and finally apply the product rule formula.

Problem 2

Find g'(x) if g(x) = (4x³ - x)(2x² + 5x - 1)

This one's a bit more complex, but the process is the same. You'll need to carefully differentiate each term in u(x) and v(x), and then meticulously apply the product rule and simplify. Remember to take your time and double-check your work!

Problem 3

Find h'(x) if h(x) = (√x + 2)(x³ - 4)

This problem introduces a square root, so you'll need to remember how to differentiate it. Rewrite √x as x^(1/2) before differentiating. Other than that, the product rule applies just as before.

Problem 4

Find k'(x) if k(x) = (x⁻¹ + 3x)(x⁴ - 2x² + 1)

Here, we have a negative exponent in x⁻¹. Remember that the power rule still applies when the exponent is negative. Be careful with your signs and take it one step at a time.

Problem 5

Find f'(x) if f(x) = (e^x)(sin x)

This problem combines the product rule with exponential and trigonometric functions. You'll need to remember the derivatives of e^x and sin x. Don't worry, it's not as scary as it looks! Just apply the product rule and the derivative rules for these functions, and you'll be fine.

Problem 6

Find g'(x) if g(x) = (x²)(ln x)

This problem involves the natural logarithm function, ln x. Remember that the derivative of ln x is 1/x. Once you've got that, the product rule is straightforward.

Problem 7

Find h'(x) if h(x) = (x^5 + 2x)(tan x)

This one includes the tangent function, tan x. The derivative of tan x is sec² x. Use this, along with the product rule, to find the derivative.

Tips for Solving These Problems

  • Write Down the Product Rule Formula: Before you start, write down the formula f'(x) = u'(x)v(x) + u(x)v'(x). This will help you keep track of the terms and avoid mistakes.
  • Identify u(x) and v(x): Clearly identify the two functions that are being multiplied together.
  • Find u'(x) and v'(x): Take your time to differentiate u(x) and v(x) correctly. Double-check your work, especially when dealing with powers, roots, or negative exponents.
  • Apply the Formula: Plug u(x), v(x), u'(x), and v'(x) into the product rule formula.
  • Simplify: Expand any products and combine like terms to get the simplest form of the derivative.
  • Check Your Work: If you have time, review your steps and make sure you haven't made any errors.

Conclusion

These practice problems are designed to help you master the product rule. Work through them carefully, and don't be afraid to make mistakes. Mistakes are a part of the learning process. The key is to learn from them and keep practicing. With enough practice, you'll become confident and skilled at applying the product rule to any derivative problem. Good luck, and happy differentiating!