Problem 13

Question

Find \(f^{\prime}(x)\). \(f(x)=\frac{x^{2}}{3 x-4}\)

Step-by-Step Solution

Verified
Answer
The derivative is \( f^{\prime}(x) = \frac{3x^2 - 8x}{(3x - 4)^2} \).
1Step 1: Identify the Differentiation Rule to Use
The function given, \( f(x) = \frac{x^2}{3x-4} \), is a rational function of the form \( \frac{u(x)}{v(x)} \). For such functions, use the Quotient Rule, which is: \( \left( \frac{u}{v} \right)^{\prime} = \frac{u^{\prime}v - uv^{\prime}}{v^2} \).
2Step 2: Determine Components of the Quotient Rule
Let \( u(x) = x^2 \) and \( v(x) = 3x - 4 \). Compute the derivatives: \( u^{\prime}(x) = 2x \) and \( v^{\prime}(x) = 3 \).
3Step 3: Apply the Quotient Rule
Substitute the expressions and their derivatives into the Quotient Rule formula:\[ f^{\prime}(x) = \frac{(2x)(3x - 4) - (x^2)(3)}{(3x - 4)^2} \].
4Step 4: Simplify the Numerator
Expand the terms in the numerator:\( (2x)(3x - 4) = 6x^2 - 8x \) and \( (x^2)(3) = 3x^2 \).Thus, the numerator becomes:\[ f^{\prime}(x) = \frac{6x^2 - 8x - 3x^2}{(3x - 4)^2} \].
5Step 5: Combine Like Terms
Combine like terms in the numerator: \( 6x^2 - 3x^2 = 3x^2 \).Thus, we have:\[ f^{\prime}(x) = \frac{3x^2 - 8x}{(3x - 4)^2} \].
6Step 6: Write the Final Derivative
The simplified expression for the derivative is:\[ f^{\prime}(x) = \frac{3x^2 - 8x}{(3x - 4)^2} \].This is the final derivative of the given function.

Key Concepts

Quotient RuleRational FunctionsDifferentiation Techniques
Quotient Rule
The Quotient Rule is an essential tool in calculus, especially when dealing with rational functions. This rule helps us find the derivative of a quotient of two functions, that is, one function divided by another. For any functions \(u(x)\) and \(v(x)\), the derivative of the quotient \( \frac{u(x)}{v(x)} \) is calculated using the formula:
  • \( \left( \frac{u}{v} \right)^{\prime} = \frac{u^{\prime}v - uv^{\prime}}{v^2} \)
This formula might seem complex at first. But it's simply saying that you take the derivative of the "top" function \(u(x)\), multiply by the "bottom" function \(v(x)\), and then subtract the product of the top function and the derivative of the bottom function. Finally, you divide the whole expression by the square of the bottom function \(v(x)^2\). Make sure that the denominator \(v(x)\) is not zero, as division by zero is undefined.
Practicing with the Quotient Rule strengthens your ability to handle more complex functions, improving your calculus skills overall.
Rational Functions
Rational functions are expressions formed by the ratio of two polynomials and appear frequently in calculus. An example is the function given in the exercise: \(f(x) = \frac{x^2}{3x-4}\). Rational functions can have a wide range of characteristics, such as asymptotes and discontinuities.
When differentiating rational functions, the numerator and denominator play key roles. Understanding how to manipulate polynomials, like expanding and combining like terms, is vital. This comes in handy when using the Quotient Rule.
Whenever you work with rational functions, keep an eye on the denominator. Ensure that it isn't zero at any point of interest, as this will make the function and its derivative undefined. Such issues are critical for determining the domain and behavior of the function.
Differentiation Techniques
Differentiation techniques are methods used to find the derivative of a function. The Quotient Rule is just one of many techniques available. Other techniques include the Power Rule, the Product Rule, and the Chain Rule, each applicable depending on the type of function.
To choose the right technique, examine the function's structure:
  • If the function is a simple polynomial like \(x^2\), the Power Rule is most suitable: \(\text{If } f(x) = x^n, f^{\prime}(x) = nx^{n-1}\).
  • If the function is a product of two functions, the Product Rule is used.
  • For composite functions, the Chain Rule is necessary.
With practice, identifying which technique to use becomes second nature. Each technique has its unique formula and understanding them well allows you to tackle complex calculus problems with confidence.