Derivatives Of Functions Using The Quotient Rule: Problems

Hey guys! Let's dive into some problems where we need to find the derivatives of functions using the quotient rule. This rule is super handy when you're dealing with functions that are fractions, where both the numerator and the denominator have 'x' in them. So, grab your pencils, and let's get started!

Understanding the Quotient Rule

Before we jump into solving problems, let's quickly recap the quotient rule. If you have a function ${y = \frac{u(x)}{v(x)}}$, where both ${u(x)}$ and ${v(x)}$ are functions of ${x}$, then the derivative ${\frac{dy}{dx}}$ is given by:

${\frac{dy}{dx} = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{[v(x)]^2}}$

Where:

• ${u'(x)}$ is the derivative of ${u(x)}$.
• ${v'(x)}$ is the derivative of ${v(x)}$.
• ${v(x)]^2}$ is the square of the original denominator.

In simpler terms, think of it like this: "(Bottom times derivative of the top) minus (top times derivative of the bottom), all over the bottom squared." Got it? Great! Let's tackle some examples.

Problem 1: ${y = \frac{1}{3x^2 + 1}}$

Okay, let's start with a classic example: ${y = \frac{1}{3x^2 + 1}}$. Here, our numerator ${u(x) = 1}$ and our denominator ${v(x) = 3x^2 + 1}$. To find the derivative, we'll follow these steps:

1. Find the derivatives of u(x) and v(x):
   • ${u(x) = 1}$, so ${u'(x) = 0}$ (because the derivative of a constant is always zero).
   • ${v(x) = 3x^2 + 1}$, so ${v'(x) = 6x}$ (using the power rule).
2. Apply the quotient rule formula: ${\frac{dy}{dx} = \frac{(3x^2 + 1) \cdot 0 - 1 \cdot 6x}{(3x^2 + 1)^2}}$
3. Simplify: ${\frac{dy}{dx} = \frac{0 - 6x}{(3x^2 + 1)^2} = \frac{-6x}{(3x^2 + 1)^2}}$

And there you have it! The derivative of ${y = \frac{1}{3x^2 + 1}}$ is ${\frac{-6x}{(3x^2 + 1)^2}}$. See? Not too scary, right?

Breaking Down the Solution

Let’s break down why each step is important. In this problem, we started by identifying the numerator and denominator as separate functions. Recognizing that the numerator, ${u(x) = 1}$, is a constant is crucial because its derivative, ${u'(x)}$, is simply 0. This often simplifies the entire expression, as we saw here. The denominator, ${v(x) = 3x^2 + 1}$, requires the power rule to differentiate, resulting in ${v'(x) = 6x}$. Once we have these derivatives, the mechanical part of applying the quotient rule formula comes into play. The substitution into the formula must be precise to ensure the correct result. Finally, simplifying the expression is not just about making it look neat; it’s about ensuring that you have the most reduced and understandable form of the derivative. This simplified form is what you’d use for any further analysis, such as finding critical points or inflection points of the function. This careful, step-by-step approach is what makes the quotient rule manageable and prevents errors.

Problem 2: ${y = \frac{2}{5x^2 - 1}}$

Next up, let's tackle ${y = \frac{2}{5x^2 - 1}}$. Similar to the first problem, we'll identify our numerator and denominator, find their derivatives, and then use the quotient rule. Ready?

1. Identify u(x) and v(x):
   • ${u(x) = 2}$
   • ${v(x) = 5x^2 - 1}$
2. Find the derivatives:
   • ${u'(x) = 0}$ (derivative of a constant)
   • ${v'(x) = 10x}$ (using the power rule)
3. Apply the quotient rule: ${\frac{dy}{dx} = \frac{(5x^2 - 1) \cdot 0 - 2 \cdot 10x}{(5x^2 - 1)^2}}$
4. Simplify: ${\frac{dy}{dx} = \frac{0 - 20x}{(5x^2 - 1)^2} = \frac{-20x}{(5x^2 - 1)^2}}$

Boom! The derivative of ${y = \frac{2}{5x^2 - 1}}$ is ${\frac{-20x}{(5x^2 - 1)^2}}$. We're on a roll, guys!

In-Depth Analysis

For the function ${y = \frac{2}{5x^2 - 1}}$, the principles remain the same but with slightly different values, offering us another chance to solidify our understanding. The numerator, ${u(x) = 2}$, is again a constant, making its derivative ${u'(x) = 0}$. This is a pattern worth noting—constant numerators will always lead to a simplified term in the quotient rule. The denominator, ${v(x) = 5x^2 - 1}$, differentiates to ${v'(x) = 10x}$, once again using the power rule. When we apply the quotient rule, the zero from ${u'(x)}$ eliminates the entire first term in the numerator, which simplifies our calculation significantly. This is a common occurrence in quotient rule problems, and spotting these simplifications early can save time and reduce errors. The key takeaway here is that recognizing constants and applying differentiation rules methodically leads to an efficient solution. Furthermore, the algebraic simplification at the end is crucial, ensuring that we present the derivative in its simplest form, ready for further applications or analysis.

Problem 3: ${y = \frac{x - 1}{x + 1}}$

Alright, let's amp things up a bit! This time we've got variables in both the numerator and the denominator: ${y = \frac{x - 1}{x + 1}}$. Don't worry, we'll tackle it step by step.

1. Identify u(x) and v(x):
   • ${u(x) = x - 1}$
   • ${v(x) = x + 1}$
2. Find the derivatives:
   • ${u'(x) = 1}$
   • ${v'(x) = 1}$
3. Apply the quotient rule: ${\frac{dy}{dx} = \frac{(x + 1) \cdot 1 - (x - 1) \cdot 1}{(x + 1)^2}}$
4. Simplify: ${\frac{dy}{dx} = \frac{x + 1 - x + 1}{(x + 1)^2} = \frac{2}{(x + 1)^2}}$

Sweet! The derivative of ${y = \frac{x - 1}{x + 1}}$ is ${\frac{2}{(x + 1)^2}}$. Notice how careful algebra is key in the simplification step. This type of algebraic manipulation is a recurring theme in calculus, so mastering it is crucial.

Deeper Dive into the Process

In this example, ${y = \frac{x - 1}{x + 1}}$, we encounter a situation where both the numerator and the denominator are linear functions. This means the derivatives of both ${u(x) = x - 1}$ and ${v(x) = x + 1}$ are constants, specifically ${u'(x) = 1}$ and ${v'(x) = 1}$. This simplifies the differentiation step but places greater emphasis on the algebraic manipulation that follows. Applying the quotient rule gives us an expression that requires careful distribution and simplification. The key here is to accurately distribute the terms and combine like terms in the numerator. Notice how the ${x}$ terms cancel out, leading to a constant numerator of 2. This kind of simplification is typical when dealing with linear functions in the quotient rule and highlights the importance of a solid algebraic foundation. The final form, ${\frac{2}{(x + 1)^2}}$, is clean and ready for any further analysis, such as finding critical points or analyzing the function’s behavior.

Problem 4: ${y = \frac{2x - 1}{x - 1}}$

Let’s keep the momentum going with ${y = \frac{2x - 1}{x - 1}}$. This one’s similar to the previous problem but with slightly different coefficients. Let's break it down:

1. Identify u(x) and v(x):
   • ${u(x) = 2x - 1}$
   • ${v(x) = x - 1}$
2. Find the derivatives:
   • ${u'(x) = 2}$
   • ${v'(x) = 1}$
3. Apply the quotient rule: ${\frac{dy}{dx} = \frac{(x - 1) \cdot 2 - (2x - 1) \cdot 1}{(x - 1)^2}}$
4. Simplify: ${\frac{dy}{dx} = \frac{2x - 2 - 2x + 1}{(x - 1)^2} = \frac{-1}{(x - 1)^2}}$

Excellent! The derivative of ${y = \frac{2x - 1}{x - 1}}$ is ${\frac{-1}{(x - 1)^2}}$. You're getting the hang of this!

Mastering the Algebraic Steps

With the function ${y = \frac{2x - 1}{x - 1}}$, we continue to reinforce the principles of the quotient rule but with a slight twist in the algebra. Differentiating ${u(x) = 2x - 1}$ yields ${u'(x) = 2}$, and differentiating ${v(x) = x - 1}$ gives ${v'(x) = 1}$. So far, so good. The application of the quotient rule leads to an expression that, like the previous problem, hinges on careful algebraic manipulation. The critical step here is the distribution of the constants and the correct handling of the negative sign in front of the term ${(2x - 1)}$. A common mistake is to incorrectly distribute this negative sign, which can lead to an incorrect derivative. Attention to detail in these algebraic steps is paramount. After correctly distributing and combining like terms, we arrive at the simplified derivative ${\frac{-1}{(x - 1)^2}}$. This exercise reinforces the idea that calculus often requires a strong command of algebraic techniques to arrive at the correct answer.

Problem 5: ${y = \frac{2x^2 - 3x + 1}{3x + 5}}$

Okay, guys, let's crank up the complexity a notch. Now we've got a quadratic in the numerator: ${y = \frac{2x^2 - 3x + 1}{3x + 5}}$. But don't sweat it, we'll still use the same trusty quotient rule.

1. Identify u(x) and v(x):
   • ${u(x) = 2x^2 - 3x + 1}$
   • ${v(x) = 3x + 5}$
2. Find the derivatives:
   • ${u'(x) = 4x - 3}$
   • ${v'(x) = 3}$
3. Apply the quotient rule: ${\frac{dy}{dx} = \frac{(3x + 5) \cdot (4x - 3) - (2x^2 - 3x + 1) \cdot 3}{(3x + 5)^2}}$
4. Simplify: ${\frac{dy}{dx} = \frac{(12x^2 + 11x - 15) - (6x^2 - 9x + 3)}{(3x + 5)^2}}$ ${\frac{dy}{dx} = \frac{12x^2 + 11x - 15 - 6x^2 + 9x - 3}{(3x + 5)^2}}$ ${\frac{dy}{dx} = \frac{6x^2 + 20x - 18}{(3x + 5)^2}}$

Nice work! The derivative of ${y = \frac{2x^2 - 3x + 1}{3x + 5}}$ is ${\frac{6x^2 + 20x - 18}{(3x + 5)^2}}$. See how expanding and simplifying polynomials becomes crucial here? These are fundamental skills that are frequently tested in calculus.

Navigating Polynomials in the Quotient Rule

The problem ${y = \frac{2x^2 - 3x + 1}{3x + 5}}$ introduces polynomials of higher degree, increasing the complexity of both the differentiation and the simplification steps. The derivative of the quadratic numerator ${u(x) = 2x^2 - 3x + 1}$ is ${u'(x) = 4x - 3}$, and the derivative of the linear denominator ${v(x) = 3x + 5}$ is ${v'(x) = 3}$. The quotient rule then requires us to multiply, distribute, and combine like terms. The expansion of the product ${(3x + 5)(4x - 3)}$ and the distribution of the ${-3}$ across ${(2x^2 - 3x + 1)}$ are critical steps where errors can easily occur. The key is to proceed methodically, ensuring that each term is multiplied correctly and that the signs are handled appropriately. After expanding and combining like terms, we arrive at the simplified derivative ${\frac{6x^2 + 20x - 18}{(3x + 5)^2}}$. This type of problem underscores the importance of not just knowing the quotient rule but also being proficient in polynomial arithmetic.

Problem 6: ${y = \frac{x^2}{x^2 + 1}}$

Last but not least, let's tackle ${y = \frac{x^2}{x^2 + 1}}$. This problem will give us a chance to really flex those quotient rule muscles.

1. Identify u(x) and v(x):
   • ${u(x) = x^2}$
   • ${v(x) = x^2 + 1}$
2. Find the derivatives:
   • ${u'(x) = 2x}$
   • ${v'(x) = 2x}$
3. Apply the quotient rule: ${\frac{dy}{dx} = \frac{(x^2 + 1) \cdot 2x - x^2 \cdot 2x}{(x^2 + 1)^2}}$
4. Simplify: ${\frac{dy}{dx} = \frac{2x^3 + 2x - 2x^3}{(x^2 + 1)^2}}$ ${\frac{dy}{dx} = \frac{2x}{(x^2 + 1)^2}}$

Fantastic! The derivative of ${y = \frac{x^2}{x^2 + 1}}$ is ${\frac{2x}{(x^2 + 1)^2}}$. We've conquered all the problems!

Strategic Simplification Techniques

The function ${y = \frac{x^2}{x^2 + 1}}$ provides an excellent example of how strategic simplification can make a significant difference. The derivatives of both the numerator ${u(x) = x^2}$ and the denominator ${v(x) = x^2 + 1}$ are straightforward: ${u'(x) = 2x}$ and ${v'(x) = 2x}$. Applying the quotient rule leads to an expression where careful distribution is crucial. However, the real key to this problem is recognizing that terms can cancel out if you distribute correctly. After expanding, we get ${2x^3 + 2x - 2x^3}$ in the numerator. The ${2x^3}$ and ${-2x^3}$ terms cancel each other out, leaving us with a much simpler expression: ${\frac{2x}{(x^2 + 1)^2}}$. This highlights the importance of looking for opportunities to simplify expressions, as it not only makes the final answer cleaner but also reduces the chance of errors in subsequent steps. Strategic simplification is a hallmark of strong calculus skills.

Conclusion

So there you have it, guys! We've tackled six different problems using the quotient rule. Remember, the key is to break down each problem into manageable steps: identify u(x) and v(x), find their derivatives, apply the quotient rule formula, and then carefully simplify. Keep practicing, and you'll become a quotient rule pro in no time! You've got this! Remember, consistent practice is the cornerstone of calculus mastery. So, keep working through problems, and don't hesitate to review and revisit concepts as needed. Calculus is a challenging but incredibly rewarding field, and with dedication and the right approach, you can conquer it!